Vaderlind P. the Quest for Functions. Functional Equations for the Beginners (2005)(46s)_MSch

8/10/2019 Vaderlind P. the Quest for Functions. Functional Equations for the Beginners (2005)(46s)_MSch

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THE QUEST FOR FUNCTIONS

Paul Vaderlind, Stockholm University

January, 2005

Functional Equations for The Beginners

Introduction ......................................................................................1

Some easy tricks ...............................................................................5

More sofisticated methods ..............................................................10

Related questions ............................................................................21

Collection of Problems ...................................................................22

Solutions to the Problems of the Collection....................................27Additional Problems .......................................................................38

INTRODUCTION

Functional equations is a rather popular topic at the IMO and other mathematical

competitions, both national and international. At least 19 IMO-problems can be

classified as functional equations and all these problems are listed below. The

question posed in this type of problems is to find all functions satysfying the given

equation and, possibly, some additional conditions like continuity, monotonicity

or being bounded.

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There however is no general method of solving this kind of problems and the

present text offers only some basic ideas that may turn out to be useful. Some type

of tricks are used when the functions considered are f : N N, other for funk-tionsf : Q

Q, and still different methods forf : R

R. Yet another approach

may be used when we know that the functions are looking for are polynomials.(Throughout this text N,Z,Q and R denotes the sets of positive integers, integers,rational numbers and real numbers respectively. An additional + sign, like R+,

means positive. N0denotes the set of all non-negative integers {0, 1, 2, 3,...}.)As an exemple of a functional equation, consider the famous Cauchys equation

f(x+ y) = f(x) +f(y), where f : R R andx, y R. This very generalequation has in fact a very limmited family of solutions as soon as one add som

extra constrain. For example, if one demands that the solution has to be a con-

tinuous function then the only solutions are the trivial ones: the linear functions

f(x) =cx, for any real constantcEven if one demands the continuty only in one sigle point x0 R, or if one asks

forfbounded in some interval(a, b) R, or monotone, then the equation has stillonly the trivial solutions. In order to find some non-trivial solutions one has to look

beyond Lebesgue measurable functions and that such pathological solutions exist

was proved by G. Hamel (in Math. Ann. 60, (1905), 459-462).

The list of all functional equation that occurred at the IMO is the following (manymore such problems has made to the IMO short-lists):

1968.5. Letfbe a real-valued function defined for all real numbers, such that forsomea >0 we have

f(x+a) =1

2+

f(x) f(x)2 for allx.Prove thatfis periodic, and give an example of such a non-constantf fora= 1.

1972.5. f(x)andg(x)are real-valued functions defined on the real line. For all xandy,f(x + y) + f(x y) = 2f(x)g(y), f is not identically zero and |f(x)| 1for allx. Prove that |g(x)| 1for allx.

1975.6. Find all polynomialsP(x, y)in two variables such that:

(1)P(tx, ty) =tnP(x, y)for some positive integern and all realt, x, y:

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(2) for all realx, y, z :P(y+z, x) +P(z+x, y) +P(x+y, z) = 0;(3)P(1, 0) = 1.

1977.6. The functionf(x)is defined on the set of positive integers and its values

are positive integers. Given thatf(n +1)> f(f(n))for alln, prove thatf(n) =nfor alln.

1981.6. The functionf(x, y)satisfies:f(0, y) =y+ 1, f(x + 1, 0) =f(x, 1)andf(x + 1, y+ 1) =f(x, f(x + 1, y))

for all non-negative integersx, y. Findf(4, 1981).

1982.1. The functionf(n) is defined on the positive integers N and takes non-

negative integer values. Moreover f(2) = 0, f(3) > 0, f(9999) = 3333 and forallm, n N: f(m+n) f(m) f(n) = 0or1. Determinef(1982).

1983.1. Find all functionsf defined on the set of positive real numbers R+ whichtake positive real values and satisfy:

f(xf(y)) =yf(x)for allx, y; andf(x) 0as x .

1986.5. Find all functionsfdefined on the non-negative real numbers and taking

non-negative real values such that: f(2) = 0, f(x)= 0 for 0 x < 2, andf(xf(y))f(y) =f(x+y)for allx, y.

1987.4. Prove that there is no functionffrom the set of non-negative integers N0into itself such thatf(f(n)) =n+ 1987for alln N0.

1988.3. A functionfis defined on the positive integers N by:

f(1) = 1, f(3) = 3, f(2n) =f(n), f(4n+ 1) = 2f(2n+ 1)

f(n), and

f(4n+ 3) = 3f(2n+ 1) 2f(n)for alln N.Determine the number of positive integersn 1988for whichf(n) =n.

1990.4. Construct a function from the set of positive rational numbers into itself

such thatf(xf(y)) =f(x)

y for allx, y.

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1992.2. Find all functionsfdefined on the set of all real numbers with real values,such thatf(x2 +f(y)) =y +f(x)2 for allx, y.

1993.5. Does there exist a functionf : N

N such that f(1) = 2, f(f(n)) =

f(n) +nfor alln N, andf(n)< f(n+ 1)for alln N?

1994.5. Let Sbe the set of all real numbers greater than1. Find all functionsf: S Ssuch thatfx+f(y) + xf(y) = y+f(x) +yf(x)for all x, y, andf(x)

x is strictly increasing on each of the intervals 1< x


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SOME EASY TRICKS

1. Transformation of variables.

This is a really basic trick and may be used as a part of a solution of a more

complex problem. Generally, given an equation of a type f(g(x)) = h(x), withg(x), h(x)given functions, then, ifg(x)has an inverse then, letting t = g(x), wegetf(x) =h(g1(x)).

Lets solve the following equation:

Example 1. Find all functions f(x) defined for all real numbers, such that f

x+ 1x

=

1 +1

x+

1

x2for allx = 0.

Solution. By lettingt=x+ 1

x , we getx=

1

t 1 . Hence, after some calculations,the equation reduces tof(t) =t2 t+ 1. Thusf(x) =x2 x+ 1.

2. Creating simultaneous equations.

This is another simple trick, which often works when the equation involves two

values f(g(x)) and f(h(x)), for two different algebraic expressions g(x) and h(x).Consider the equation:

Example 2. Find all functionsf : R

R such that x2f(x) + f(1

x) = 2x

x4

for allx R.Solution.Replacingxby1 x, we have(1 x)2f(1 x) +f(x) = 2(1 x) (1 x)4. Sincef(1 x) = 2x x4 x2f(x)by the given equation, substitutingthis into the last equation and solving forf(x), we getf(x) = 1 x2.

Now we should check that this function satisfiy the given equation: x2f(x) +f(1 x) =x2(1 x2) + 1 (1 x)2 = 2x x4.

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One more example:

Example 3. Solve the equationf1

x

+

1

xf(x) =x, wherefis a real valued

function defined for all real numbers except 0.

Solution.Replacingx by 1

x yeldsf(x) +xf

1

x

=

1

x. Replacing nowxby

xleads to a new equationf(x) xf1

x

=1

x. From this equation and the

original one can we now find the functionf(x): Multiply the first equation withx

and add to the last one. What we get is2f(x) = x2 1x

. Replacing once again

x by

x we get the final answer: f(x) =

x3 + 1

2x

. It remains to verify that this

function satisfies the given eqation.

Remark. In most cases we solve the equation under the (silent) assumption that

the functionf(x)exists. As a consequence, it is necessary to check that the obtai-ned function really satisfies the given equation.

3. Using symmetry.

If possible, one should use symmetry when dealing with the equation involving

more than one variable.

Example 4. Find all functionsf : R R, such thatf(x+y) =x+f(y)for allx, y R.

Solution. Left-hand side of the equation is symmetric inxandy. Thusx + f(y) =f(x+y) =f(y+x) =y +f(x), which can be written asf(x) x= f(y) y,for all x, y R. Hence f(x) x is constant for all x R, and the answer isf(x) =x+c, for any choice of the real constant c, provided that those functionssatisfy the given equation. This however can be easily checked.

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Example 5. Find all functionsf : R R, such thatf(x+y) f(x y) = 4xyfor allx, y R.

Solution. Let u = x+ y andv = x y. Then the equation can be written asf(u) f(v) = u

2

v2

, or f(u) u2

= f(v) v2

. Since this relation holdsfor arbitraryu, v R thenf(u) u2 is constant. Thus, f(u) u2 = c and theanswer is the family of functionsf(x) =x2+c, for any choice of a real constantc,provided these functions satisfy the given equation. This however is easy (although

necessary) to check.

4. Evaluatingf(x0)for some special choices ofx0.

Findingf(x0)for some values ofx0, likef(0),f(1),f(2),f(1)and so on, maygive some ideas on the structure off(x). This seems to be specially usefull whenthe equation involves more than one variable.

Example 6. (Korea, 1988) Findf : R R, such thatf(x)f(y) =f(xy) + x + yfor allx, y R.

Solution. Letting y = 0 we get f(x)f(0) = f(0) +x. Hence, f(0)= 0 andf(x) =

x

f(0)+ 1. Taking nowx = 0we find thatf(0) = 1. Thusf(x) = x+ 1

and it is easy to verify that this function satisfies the given eqation.

In the next example, let Q+ denote the set of positive rational numbers.

Example 7. Find all functions f : Q+ Q+, such thatfx+ yx = f(x) +f(y)

f(x)+ 2y, for allx, y Q+.

Solution. By letting (x, y) be (1, 1), (1, 2) and (2, 2) we find out that f(2) =

f(1) + 3, f(3) = f(1) + f(2)

f(1) + 4 and f(3) = f(2) + 5. From these three

equalities we can deduce thatf(1) = 1, f(2) = 4and f(3) = 9. This leads to the

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hypothesis thatf(n) =n2, for at leastn N.This hypothesis may be now verified by taking x = y = n and using the

obtained relation f(n + 1) = f(n) + 1 + 2n together with the mathematicalinduction.

We may now suspect that the only solution of the equation is the functionf(x) = x2, for all x Q+. Lets take first x = n, y = m and then x = m

n,

y=m (forn, m N). We getf

n+m

n

=f(n) +

f(m)

f(n) + 2m= n2 +

m2

n2 + 2m and

fm

n +n

=f

mn

+

f(m)

fmn

+ 2m= fmn

+

m2

fmn

+ 2m.From the last two equalities it follows thatn2 +

m2

n2 =fm

n+ m2

fmn , which

can be expressed as

0 = fm

n

m2n2

n2 + m2

fmn

= fmn

m2n2

n2

fmn

fmn

m2n2

=

fm

n

m2n2

1 n

2

fmn

.Let

p

g Q+, wherep, q N. If1 q

2

fpq= 0then, according to the equality

above,fpq

p2q2

= 0, i.e.fpq

= p

2

q2 =pq

2.

If1 q2

fpq

= 0 then, f(2q)f2p2q

= 4q2fpq

= q2fpq

= 1. Thus f(2q)f2p2q

= 1,and then, letting n = 2q, m = 2p into the equality above, we find again that

fp

q

=f

2p2q

=

(2p)2

(2q)2 =p

q

2.

Hence the answer isf(x) =x2, and it is easy to verify that this function satis-

fies the equation.

5. Polynomials.

When the functions we are looking for are polynomials there are several several

properties one should take into the consideration. The most importatnt are: the

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degree, the finite number of zeroes (unless the polynomial is the trivial one:p(x) 0) and the Factor Theorem (stating that p() = 0if and only ifx is a divisorofp(x)).

Example 8. Find all real polynomialsp(x) such thatp(x+1)+2p(x1) = 6x2+5for allx R.

Solution. First we observe that p(x) has to be of degree 2, hence we may writep(x) = ax2 +bx+c. Substituting this expresion into the equation we get a(x+1)2 +b(x+ 1) +c+ 2a(x 1)2 + 2b(x 1) + 2c = 6x2 + 5, which reduces to3ax2 + (2a + 3b)x + (3a b + 3c) = 6x2 + 5. Identifying the coefficients givesa = 2, b =

4

3, c =

1

9. Hence p(x) = 2x2 +

4

3x+

1

9and verification that this

polynomial satisfies the given relation is an easy task.

Example 9. Find all real polynomialsp(x)such thatxp(x 1) = (x 2)p(x)forallx R.

Solution.Lettingx = 0we get0 =2p(0), i.e.p(0) = 0. Similarily, forx = 2we getp(1) = 0. Hencep(x) is divisible byx and by(x 1)and we can write

p(x) =x(x 1)q(x), whereq(x)is a polynomial of degree 2 less than the degreeofp(x).Replacingp(x)withx(x 1)q(x)in the original equation givesx(x 1)(x

2)q(x 1) = (x 2)x(x 1)q(x)for allxR. Henceq(x 1) =q(x)for allx R.

Let now x0 be any fixed real number and consider the polynomial h(x) =q(x)q(x0). It is obvious that h(x0) = 0. Moreover, h(x0 + 1) = q(x0 +1)q(x0) = q(x0)q(x0) = 0, and, usig the induction, one can show thath(x0+n) = 0for alln

Z.

Since a non-zero polynomial only has a finite number of zeroes, then h(x) 0.It implies thatq(x)is a constant polynomial, and thenp(x) =cx(x 1), for anychoice of a real constantc.

It only remains to check that the polynomials p(x) = cx(x 1) satisfy theoriginal equation.

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MORE SOFISTICATED METHODS

8. Continous functions.

Some equations involving continous functions f : R R may be solved in thefollowing way: Find first some special values, like f(0) or f(1). By inductiondetermine then the values f(n)for allnN, followed by the valuesf(n)for alln Z. In the next step find the valuesf1

n

forn Z and then findfm

n

for

allm

n Q. Finally, use thecontinuityoff(x)and the fact that the set of rational

numbers isdensein R, to determaine the formula for f(x)for allx R.That the the set of rational numbers Q is dense in R means that for eachx R

there exists a sequence {xn} of rational numbers such that limn

xn=x.

Suppose a function f(x) is defined on the subset I R Then we say thatf(x) is continuous at a point x0 I if, for each sequence{xn} I such thatlimn

xn=x0, we have limn

f(xn) =f( limn

xn) =f(x0).

We say thatf(x)is continous onIif it is continous at each pointx0 I.As an illustration consider the already mentioned continous version of Cauchys

equation:

Example 10. Find all continous functionsf : R R such that f(x+ y) =f(x) +f(y)for allx, y R.

Solution. Lettingx = y = 0 into the equation we get f(0) = 0. By inductionone shows easily that f(nx) = nf(x), for all n N and all x R. Hencef(n) =nf(1), for alln

N.

If we in the equation let y =x then we get f(0) = f(x) +f(x). Thusf(x) =f(x) for all x R. For n N we have then f(n) =f(n) =nf(1), which means thatf(n) =nf(1)is valid for alln Z.

Suppose now thatm N andn Z. Thenn fmn

=f

n m

n

=f(m) =

mf(1). Thusfm

n

=

m

nf(1), i.e.f(x) =xf(1)is valid for allx Q.

Suppose finally that x R but x / Q. Then, since Q is dense in R, con-

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sider a sequence{xn} od rational numbers such that limn

xn = x. By the con-

tinuity off(x) we have f(x) = f( limn

xn) = limn

f(xn) = limn

xnf(1)

=

f(1) limn

(xn) =f(1) x.

Hencef(x) = xf(1)for allx R. Since f(1)can be any real number thenthe solution, if exists, must be of the form f(x) = cxfor any real constantc. Asusual, it remains to verify that these functions satisfy the original equation.

Example 11. Find all continous functionsf(x)defined for x > 0 and such that

f(x+y) =f(x)f(y)

f(x)f(y)for allx, y R+.

Solution. It is obvious that f(x) = 0 for all x R+

. Taking x= y we get f(2x) =f(x)f(x)

f(x) +f(x) =

1

2f(x). For y = 2x we have then f(3x) =

f(x)f(2x)

f(x) +f(2x) =

f(x)12

f(x)

f(x) + 12f(x) =

1

3f(x). This suggest that f(nx) =

1

nf(x) for all n N, and

may easily be shown by induction.

By taking x = 1 in the last equality we get f(n) = 1

nf(1) for all n N.

Moreover,f(1) = fn 1

n = 1

n

f1n, which means that f1

n = nf(1). Then,

for allm, n N, fmn

=f

m 1

n

=

1

mf1

n

=

n

mf(1).

We have that far shown thatf(x) = 1

xf(1)for allx Q+. Using the continuity

argument this can be extended to all x R+. The answer is thenf(x) = cx

for

every non-zero real constantc.

Example 12. (Croatia, 1996) Supposet is a fixed number such that0 < t < 1.Find all functions f : R R, continous at x = 0, such that f(x) 2f(tx) +f(t2x) =x2 for allx R.

Solution. Since the equation can be written as

f(x)f(tx)f(tx)f(t2x) =x2, we may start by a substitution g(x) = f(x) f(tx). This will simplify theequation tog(x) g(tx) =x2.

Now, sincef(x)is contious at x = 0, then is obvious that even g(x)is conti-

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nous atx= 0and thatg(0) =f(0) f(0) = 0.In the equation g(x) g(tx) = x2 we now sibstitute x bytx several times,

getting successively:

g(x)

g(tx) =x2,

g(tx) g(t2x) =t2x2,g(t2x) g(t3x) =t4x2,.......

g(tn1x) g(tnx) =t2(n1)x2.Adding all thos equalities we find thatg(x) g(tnx) = 1 +t2 +t4 +...+

t2(n1)

x2, and, since t2 = 1 then g(x)g(tnx) = x21 t2n

1 t2 = x2 1

1 t2

x2 t2n

1 t2

.

Remembering that0 < t < 1we can now let n . Thentnx 0as wellas

t2n

1 t2 0and, using the continuity ofg(x)at x = 0, we getg(x) g(0) =

x2 1

1 t2 . Sinceg(0) = 0then finallyg(x) = x2

1 t2 .

We have that far found out that f(x) f(tx) = x2

1 t2 for all x R. Whatwe can do now is to repeat the same procedure we did above: substitution xbytx

several times. We get:f(x) f(tx) = x

2

1 t2 ,

f(tx) f(t2x) = t2x2

1 t2 ,

f(t2x) f(t3x) = t4x2

1 t2 ,.......

g(tn1x) g(tnx) = t2(n1)x2

1

t2 .

Adding those equations we find out thatf(x) f(tnx) = x21 t2

1 + t2 + t4 +

...+t2(n1)

= x2

1 t21 t2n1 t2 .

Letting nown and using the continuity off(x)at x = 0, we getf(x) f(0) =

x2

(1 t2)2 for allx R.

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x+ 4instead ofx yeldsf(x+ 12) +f(x) = 0. Since thenf(x+ 24) +f(x+12) = f

(x+ 12) + 12

+f(x+ 12) = 0 the the last two equations impliy that

f(x+ 24) =f(x)for allx R.Thus we have found a common period p= 24for allf(x)satisfying the origi-

nal equation. Now the question is if this period is the least positive one.Consider the function f(x) = sin

x

12. Since 2 is the least positive period

of sin x then p = 24 is the least positive period off(x). At the same time it

is easy to show that f(x) satisfies the condition of the problem:sin(x+ 4)

12 +

sin(x 4)

12 = sin

x12

+

3

+ sin

x12

3

, which, by easy trigonometry,

reduces tosinx

12.

Hencet the least common period isp= 24.

Example 14. (Romania, 1999) Suppose that the functionf : N N is surjective,while the functiong : N N is injective. Given thatf(n) g(n)for alln N,prove thatf=g.

Solution. LetA= {n N : f(n) =g(n)} and supposeAis a non-empty subsetofN. Then the set B ={g(n) : n A} is also a non-empty subset ofN andthus has the least element. Suppose g(a), for someaA, is the least element ofb. Then, sinceg(n)is injective, we haveg(a) < g(b), for alla= b A and, bythe definition ofA,g(a)< f(a).

Sincef(n)is surjective then there existsc N such thatf(c) =g(a)< f(a).Note that c= a. Now, since g(n) is injective then g(c)= g(a) = f(c). Hencec Aand we haveg(c)< f(c) =g(a), which contradicts the choice ofa. Thus,the setA is empty, which means thetf(n) =g(n)for alln N,

Example 15. (IMO, 1983) Find all functionsf : R+ R+ such thatf(xf(y)) =yf(x)for allx, y R+, andf(x) 0asx .

Solution. By taking x = y = 1 we get f(f(1)) = f(1). Taking x = 1 and

y=f(1)yeldsf(f(f(1))) =

f(1)2

. Combining these two equalities we receivef(1)

2= f(f(f(1))) = f(f(1)) = f(1). Hence f(1)

f(1) 1 = 0. Since

f(1)> 0, then we must havef(1) = 1, i.e.x= 1is a fixed point off(x).

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Taking y = x yelds f(xf(x)) = xf(x), which means that xf(x) are fixedpoints offfor allx R+.

Suppose thatf(x)has a fixed pointx0>1. Then, by the above,x0f(x0) =x20

is a fixed point as well. Then again, x20f(x20) = x

40 is a fixed point off, and, by

induction,x2k0 are fixed points offfor allk N. Sincex0> 1 then limk x2ko = and it follows that lim

kf

x2ko

= limk

x2ko =, which contradicts the conditionstated in the problem. Thusf(x)has no fixed points greater than 1.

Lets now check iff(x)has some fixed points within the interval(0, 1). Ifx0is

such a point then, takingy =x0 andx = 1

x0into the relation we get 1 =f(1) =

f

1

x0 x0

= f

1

x0f(x0)

= x0f

1

x0, i.e.f

1

x0=

1

x0. Thus

1

x0> 1is a fixed

point off(x), which contradicts the previous result. Hencex= 1is the only fixedpoint off(x).

We have however fund earlier thatxf(x)are fixed points offfor allxR+.Thus xf(x) = 1 for all x R+, which means thatf(x) = 1

x. It is now easy to

check that this function satisy the given conditions.

Example 16. (IMO, 1987). Prove that there is no functionf : N0 N0such thatf(f(n)) =n+ 1987for alln

N0.

Solution. Suppose there is such a function f(x). Then f(x) must be injective(one-to-one) becausef(a) =f(b)would implya= f(f(a)) 1987 =f(f(b)) 1987 = b. Moreover, it is clear that the function f(f(n)) = n+ 1987 will neverhave the values from the set {0, 1, 2, 3,..., 1986}, and those 1987 numbers are theonly one from N0 that the functionf(f(n))will miss ().

Suppose now thatf(n)misses exactlyk distinct valuesc1, c2,...,ck in N0, i.e.f(n)

=c1, c2,...,ckfor alln

N0. This implies thatf(f(x))misses the following

2k values: c1, c2,...,ck, f(c1), f(c2),...,f(ck) in N0. (Note that all the numbersf(cj)are distinct, sincefis injective.)

Now, ifw / {c1, c2,...,ck, f(c1), f(c2),...,f(ck)}, then there is m N0 suchthatf(m) =w. Sincew =f(c1), f(c2),...,f(ck)andm =c1, c2,...,ck so there isn N0such thatf(n) =m. Hencef(f(n)) =w.

This proves that the functionf(f(n))misses only the2k values {c1, c2,...,ck,f(c1), f(c2),...,f(ck)} and no others. This contradicts the fact stated as () above

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(1987 is an odd number).

Example 17. (IMO, 1968). Letf(x)be a real-valued function defined for all realnumbers

x, such that for some positive constant

athe equation

f(x+a) =1

2+

f(x) f(x)2 holds for allx R.

Prove thatf(x)is periodic, and, fora= 1, give an example of such a non-constantfunctionf(x).

Solution. One way of solving this problem is to rewrite the equation asf(x+

a) 12

=

f(x)

1 f(x), and to realize that both sides of the equality aresymmetrical about

1

2. Then it seems natural to make the substitution

g(x) =f(x) 1

2.

With this substitution we will haveg(x) 0and

g(x+a)2

=1

4

g(x)2

for all x. It follows that that

g(x+ 2a)2

= 1

4

g(x+ a)2

= 1

41

4

g(x)2

=

g(x)2

. Thus g(x+ 2a) =g(x) for allx.

Hence,f(x+ 2a) = g(x+ 2a) +

1

2 = g(x) +

1

2 = f(x) sof(x) is periodicwith the period2a.

There are several examples of non-constant functions satisfying the given equa-

tion and having period = 2. One such eaxample is f(x) = 1

2

1 +

cosx2

(check this!!). An another example one can get by takingf(x) to be arbitrary in

the interval[0, 1) (for example, let f(x) = 1for0 x < 1), then letf(x) = 12

for1 x < 2. Finally use the equality f(x+ 2) = f(x) to extend f(x) to allother values ofx

R.

Example 18. (IMO, 1996). Find all functionsf : N0 N0 such thatf

m+f(n)

=f

f(m)

+f(n)for allm, n N0.

Solution. Takingm = n = 0, we getf

f(0)

=f

f(0)

+f(0), which impliesthatf(0) = 0. Takingm = 0. we getf

f(n)

= f(n), i.ef(n) is a fixed point

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off(x)for alln N. As a consequence, the equation becomesfm+f(n) =f(m) +f(n) ().

Now we will show by induction that ifn0 is a fixed point off(x) thenkn0 isalso a fixed point off(x) for all k

N0. We know this already for k = 0 and

k = 1. If we assume that kn0 is a fixed point off(x) for some k N0 thenf

(k + 1)n0

=f

kn0 + n0

=f

kn0 + f(n0)

=f(kn0) + f(n0) =kn0 + n0=(k+ 1)n0, and so is(k+ 1)n0 also a fixed point off(x).

If0 is the only fixed point off(x) then, by the relation (),f(m) = 0for allm N0.

Otherwisef(x)has a least fixed pointn0= 0(the least element in the set of allnon-zero fixed points off(x)). We want to show now that kn0 are the only fixedpoints off(x)(fork N0).

So suppose that x is a fixed point. Then x n0 and dividingx nyn0 we getx= kn0+r, where0 r < n0. Thusx= f(x) =f(r +kn0) =f

r +f(kn0)

=f(r) +f(kn0) =f(r) +kn0. From this it follows thatf(r) =x kn0=r. Thismeans thatr is a fixed point off(x) and by the minimality ofn0, it follows thatr= 0. Hencex= kn0 and we are done.

We have however shown that f(n) are fixed points off(x) for all n N0.Hence f(n) = cnn0 for some numbers cn N0. However c0 = 0 since 0 =f(0) =c0n0.

Dividing now each nN0 byn0 we get n = kn0+r, where0

r < n0.

Thenf(n) = f(r+ kn0) = f

r+ f(kn0)

= f(r) + f(kn0) = f(r) + kn0 =

crn0+ kn0= (cr+ k)n0=

cr+ nno

, where x denotes the integer part ofx.Hence the answer isf(n) =

cr+ n

no

, but this, of course, must be veryfied.

To this end, for eachn0>0 letc0= 0and letc1, c2,...,cn01 N0be arbitrary.The functionf(n) =

cr+ n

no

, wherer is the remainder ofn divided byn0,

are all solutions: Write m = kn0+r and n = ln0 +s, with 0 r, s < n0.Thenf

m+ f(n)

= f

r+ kn0+ (cs+ l)n0

= crn0+kn0+csn0+ln0 =

f

f(m)

+f(n). Thatf(n) 0also is a solution is obvious.

10. A good guess.

Sometimes a good guess may simplify the work considerably. There are equations

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indicates which property is being used):

g(x) (5)= g(x+ 1)

(6)= (x+ 1)2 g 1

x+ 1

(4)=(x+ 1)2 g 1

x+ 1

(5)=(x+ 1)2

g 1x+ 1

+ 1 =(x+ 1)2

g x

x+ 1 (6)=(x+ 1)

2

x2

(x+ 1)2gx+ 1

x =

x2 g1 +1x

(5)=x2 g1

x

(6)=x2 1

x2 g(x) = g(x).

Hence2g(x) = 0, i.e.g(x) = 0for allx R, andf(x) =xis the only solutionto the equation.

Example 21. Show that there are infinitely man functionsf : N N such thatf(2) = 2andf(mn) =f(m)f(n)for allm, n

N.

Solution. Eachn >1 has a unique representation as a product of prime numbers,n = pn11 p

n22 ...p

nkk , where pi are prime numbers andni N. The condition of the

problem implies then thatf(n) =

f(p1)n1f(p2)n2...f(pk)nk (). Hence the

function is defined by its values on the set of prime numbers, which may be then

choosen arbitrarily.

To exhibit one specific infinite family of solutions let P ={q1, q2, q3, ....} bethe set of all prime numbers greater that 2, in increasing order. For eachm N,let the function fm be defined onPin the following way: fm(qi) = qi+m. Thenwe may add fm(1) = 1 andfm(2) = 2, and, using the property () extend thedefinition offmto the whole N.

11. Some useful facts.

We have already worked out one of the Cauchys equations but there are another

three. Since all they are already a folklore, the complet solutions are not given here,

only the final answer, and they may be used in the solutions of other problems as

given facts. However one should recommend that the reader try to solve those

equations on his own.

Note that we only mention the continous solutions. The general cases of Cauchys

equations (i.e. without any extra conditions) are much harder to solve.

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The only continuous solutions to the following Cauchys equations:

(1) f(x+y) =f(x)f(y) for allx, y R,(2) f(xy) =f(x) +f(y) for allx, y R+,(3) f(xy) =f(x)f(y) for allx, y R+

are the following families of functions:

(1) f(x) =cx for any real constantc >0, orf(x) 0,(2) f(x) =c ln xfor any real constantc,

(3) f(x) =xc for any real constantc, orf(x) 0.

Example 22. (Example 10 revisited). Find all continous functionsf(x) defined

forx >0 and such thatf(x+y) =f(x)f(y)

f(x)f(y)for allx, y R+.

Solution. We note that f(x)= 0 for all x R+ and put g(x) = 1f(x)

. The

the equation may be written asg(x+y) = g(x) + g(y). This is the well knownCauchys equation and the continous solutions are g(x) = cx for any choice of

real constantc. Hence the solutions of the original equation aref(x) = 1

cx for allnon-zero real constantsc.

Example 23. Find all continous functionsf : R+ R+ such that fxy =f(x)f(y) for allx, y R+.

Solution. We note that the constant function f0(x) 1is a solution to the equa-tion. Suppose then that there is another solution,f(x), and thatf(a)

= 1for some

a R+. Then, for allx, y R+f(a)f(xy) =f

axy

=f

axy

=f

axf(y)

=f(a)f(x)f(y),

from which follows that f(xy) = f(x)f(y) for allx, y R+. Since this is oneof the Cauchys equations (equation (3) above), we have f(x) =xc for some realconstantc = 0.

Putting this function into the original equation it is easy to find that c = 1.

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verified by induction.

Hencef(1996) = 998.

Example 26. (Greece, 1997) Letf :

R+

R be a function satisfying following

conditions:

(1) f(x)is strictly increasing,

(2) f(x)> 1x

and

(3) f(x)f

f(x) +1

x

= 1 for allx R+.

Findf(1).

Solution. Letf(1) =a. Settingx = 1in (3) we getaf(a+ 1) = 1. Thusa

= 0

andf(a+ 1) =1a

.

Taking now x = a + 1 in (3) yelds 1

af1

a +

1

a+ 1

= 1, which implies

thatf1

a+

1

a+ 1

= a = f(1). Sincef(x) is strictly increasing, we must have

1

a+

1

a+ 1= 1. By solving this equation we get a =

1 52

.

Suppose thata = 1 +

5

2 . Then1 < a = f(1) < f(a+ 1) =

1

a < 1. This

contracition implies thatf(1) =a =1 5

2 .

(One may note that a function with described condition really exists, for ex-

amplef(x) =1 5

2x .)

COLLECTION OF PROBLEMS

The problems below are the first set of problems for training in solving functional

equations. To each problem there is given a hint, but it is not necessary to follow

it in order to find the solution. There, as almost always, are many different ways

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to approach a mathematical problem. The suggested complet solutions are given

in the next section.

1. Find all solutionsf(x)

of the equation xf(x) + 2xf(x) = 1

wherex

R

andx = 0.(Hint: Create an additional equation.)

2. Find all functions f(x) soving the equation f(x) +f 1

1 x

= x, where

x = 0andx = 1.(Hint: Create an additional equation.)

3. Solve the functional equation 2f(tan x) +f( tan x) = sin 2x, wheref(x)are definded in the interval(

2,

2).

(Hint: Transformation of variable.)

4. (Poland, 1989) Determine all functionsf : R R, such that for allx, y R,(x y)f(x+y) (x+y)f(x y) = 4xy(x2 y2).

(Hint: Similar to Example 4.)

5. Find all polynomialsp(x)satisfying the relation p(x+ 1) =p(x) + 2x+ 1.

(Hint: Discover symmetry.)

6. (Sweden, 1995) Find all polynomialsp(x)which solve the following equationfor allx R: xp(x 1) = (x 26)p(x).

(Hint: The same method as in Example 8.)

7. Determine all continous functionsf : R

R such thatf(1) = 2andf(xy) =

f(x)f(y) f(x+y) + 1 for allx, y R.(Hint: Find first the expression forf(x)forx Q.)

8. (Canada, 2002) Find all functionsf : N0 N0 such thatxf(y) + yf(x) =(x+y)f(x2 +y2)for allx, y N0.

(Hint: Try some values ofxand guess the solution. Then prove the correctnessof your guess.)

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9. (Asian-Pacific MO, 2002) Find all functions f : R R such that f(x) = 0 hasonly a finite number of roots andf(x4 +y) =x3f(x) +f

f(y)

for allx, y R.

(Hint: Show first thatf(x4) =x3f(x)for allx, y R. Prove then thatf(x)isan odd function. What are the zeros off(x)?)

10. (UK, 1977) Letf : N N0 satisfy(a) f(mn) =f(m) +f(n), for allm, n N,(b) f(n) = 0whenever the units digit ofn(in base 10) is a 3, and(c) f(10)=0.

Prove thatf(n) = 0for alln N.(Hint: Factorization.)

11. Find all functionsf(x, y)from the set Q+Q+ of allpairsof positive rationalnumbers(x, y)to the set Q+, which satisfy the following conditions:

(1) f(x, 1) =x for allx Q+,(2) f(x, x) = 1 for allx Q+ and(3) f(x, y) f(z, t) =f(xz,yt) for allx, y, z,t Q+.(Hint: No need for that. Its a very easy problem.)

12. Determine all continous functionsf : R+

R+ such thatf2(x) = f(x+

y)f(x y)for allx, y R.(Hint: To get rid of the square, take logarithms on both sides.)

13. Find all functions f : N N which satisfy the equation f

f

f(n)

+

f

f(n)

+f(n) = 3n, for alln N.(Hint: Show first thatf(n)must be injective. What isf(1)?)

14. Find all real polynomialsp(x) satisfyingp(x2

) +p(x)p(x+ 1) = 0 for allx R.(Hint: Show that ifx0is a zero of the polynomialp(x)then evenx

20is a zero of

this polynomial.)

15. (Sweden, 1962) Determine all functionsf : R R such that for all x Rand allr Q the inequality |f(x) f(r)| 7(x r)2 is satisfied.

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(Hint: Find first the values odf(x)for rationalx. The triangle-inequality maybe useful.)

16. (Israel, 1995) Leta be a real number. Determine all functions f : R+ R+

such thatax2f1

x

+f(x) = x

x+ 1 for allx R+.

(Hint: Put 1

x instead ofx and symplify the equation. Consider several cases

depending ona.)

17. (Korea, 1999) Determine all functions f : R R such that fx 3x+ 1

+

f3 +x

1

x =x for allx R,x = 1andx = 1.

(Hint: Take firsty = x 3x+ 1

and theny = 3 +x1 x . This will give two equations

which are not difficult to solve.)

18. Find all functionsf : N N such that ff(m) +f(n) = m+n for allm, n N.

(Hint: Show thatf(nis injective (one-to-one). Findf(1).)

19. (Poland, 1992) Determine all functionsf :Q+

Q+

such thatf(x+ 1) =f(x) + 1 and f(x3) =

f(x)3

for allx Q+.(Hint: Consider the rational numberx=

m

n +n2 form, n N.)

20. (Belarus, 1995) Find all functions f : R R such that ff(x+ y) =f(x+y) +f(x)f(y) xyfor allx, y R.

(Hint: This is a tricky one. One way of doing it is to try to get rid of the double

fon the left hand side. You may first put y = 0and then replacex by x+y. Trythe same trick with the new equation, but withy = 1this time.)

21. (IMO, 1982) The functionf(n)is defined on the positive integers N and takesnon-negative integer values. Moreoverf(2) = 0, f(3)> 0, f(9999) = 3333andfor allm, n N: f(m+n) f(m) f(n) = 0or1. Determinef(1982).

(Hint: Since the condition f(m+n)f(m)f(n) = 0 or 1 is not easy to handlewe may try to replace it with a (weaker) condition f(m + n) f(m) + f(n). Find

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f(3)and thenf(3n).)

22. Let f : N N be a strictly increasing function satisfying f(2) = 2 andf(mn) = f(m)f(n)for allm, n N such that(m, n) = 1. (The notion(m, n)means the greatest common divisor ofm andn. Thus,(m, n) = 1 means thatmandnare coprime.)

Prove thatf(n) =n for alln N.(Hint: Show that ifm is an odd integer and f(m) = m, thenf(2m) = 2m.

What isf(3)? Then, the indirect proof may be an effective method.)

23. (Chech Republic and Slovakia, 1993) Determine all functionsf : Z Zsuch that f(1) = f(1) and f(x) +f(y) = f(x+ 2xy) +f(y 2xy) for all

x, yZ

.(Hint: Find f(3) andf(5) in terms off(1). What pattern do you see? Show

then thatf(n)is even. What can you find about the valuef(mn)for oddm?)

24. (IMO, 1977) The functionf(x)is defined on the set of positive integers andits values are positive integers. Given thatf(n + 1)> f(f(n))for alln, prove thatf(n) =n for alln N.

(Hint: Since we are given an inequality, it may turn out to be smart to stick to

the inequalities and work on showing that f(n) nandf(n) n. Consider aswell proving the following statement: Ifm nthenf(m) n. This may be doneby induction.)

25. Solve the same problem as in Example 19, but without the assumption that

f(x)must be continous.

(Hint: After beginning as in Example 19, show that f(x+y) = f(x) + f(y)for non-constant solutionf(x). Prove then thatf(x) =x for allx Q+ and findthen a way to extend the result to all x

R+.)

26. (IMO, 2002) Find all functionsf : R R such thatf(x) +f(y)

(f(u) +f(v)

=f(xu yv) +f(xv+yu)for allx, y,u, v R.

(Hint: Findf(x)for rationalxand then try to extend the result tox R. Sincein the problem nothing is said about the continuity off(x), so you cannot use thestandard argument. Instead you may find it useful to prove that f(x) is even andmonotone forx 0.)

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SOLUTIONS TO THE PROBLEMS OF THE COLLECTION

1. First we note thatx= 0must be excluded from the domain off(X). Substitu-tingxby xyeldsxf(x) 2xf(x) = 1. Adding this equation twice to theoriginal equation gives 3xf(x) =3. Hence f(x) = 1

x. It remains to verify

that this function satisfy the given equation.

2. Replacingx by 1

1 x yelds the equqtion f 1

1 x

+fx 1

x

=

1

1 x .

Replacingx again by 1

1 x givesfx 1

x +f(x) =

x 1x

. Subtracting from

this equation the previous one and adding the original equation yields 2f(x) =x3 +x 1x(1 x) . Thus f(x) =

x3 x+ 12x(x 1) and it is easy to verify that this function

satisfy the original equation.

3. Let y = tan x. Then sin2x = 2y

y2 + 1 and the equation can be written as

2f(y) +f(y) = 2yy2 + 1

.

Replacing nowy with y gives a new equation 2f(y) + f(y) = 2y

y2 + 1 .If we now from this equation twice substract the first equation we get 3f(y) = 6y

y2 + 1, i.e.f(y) =

2y

y2 + 1. Thus,f(x) =

2x

x2 + 1, and what remains is to check

that this function satisfies the original equation.

4. Letu = x+y andv = x y. Then the equation can be written as vf(u) uf(v) = uv(u2 v2). Foru= 0 and v= 0this can be written as f(u)

u u2 =

f(v)v

v2.

Since this relation holds for arbitrary non-zero u, v R then f(u)u

u2 is

constant. Thus,f(u)

u u2 =c, i.e.f(x) =x3 +cx.

Observe that from the relationvf(u)uf(v) =uv(u2v2)follows (by takingu= 0, v= 0) thatf(0) = 0. Since for each functionf(x) =x3 + cxwe have also

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f(0) = 0then, if the original equation has a solution, it must be f(x) =x3 +cxfor any real constantc. Thus it only remains to check that this functions satisfy thegiven equation.

5. The equation can be written as p(x + 1) (x + 1)2

=p(x) x2

, or, by lettingq(x) = p(x) x2, as q(x+ 1) = q(x). By induction one can show now thatq(x+n) =q(x)for alln Z.

If we let h(x) = q(x) q(o), then it follows that h(n) = 0 for all n Z.Sinceh(x)is a polynomial then h(x) 0andq(x)is a constant polynomial. Thisimplies thatp(x) =x2 +c, for any choice of a real constant c.

Substitutingp(x)in the original equation verifies that this family of polynomi-als satisfy the equation.

6. First one should find out thatp(0) = 0(by takingx = 0) and then thatp(k 1) = 0impliesp(k) = 0for k = 1, 2,..., 25. Hencep(x) =

25k=0

(x k) q(x), forsome polynomialq(x). Letting this expression forp(x)into the original equaition

yeldsx25k=0

(x 1 k) q(x 1) = (x 26)25k=0

(x k) q(x).It follows that for x > 26 we have q(x 1) = q(x), and then, by the same

argument as in Example 8,q(x) =constant. Finallyp(x) =c25k=0

(x k), for anychoice of the real constant c. It is easy to check that these polynomials satisfy theoriginal equation.

7. In order to simplify the calculation lets introduce a new function g(x) =f(x) 1. (This is a smart substitution allowing us to get rid of the constant 1inthe relation given in the problem). Replacing thenf(x)byg(x) + 1in the relation

gives () : g(xy) +g(x+y) =g(x)g(y) +g(x) +g(y)andg(1) = 1.Insertingy = 1 in () yeldsg(x) +g(x+ 1) = g(x)g(1) +g(x) +g(1), i.e.g(x + 1) =g(x) + 1. Thus, forx= 0we haveg(1) =g(0)+1, which means thatg(0) = 0. Moreover, takingx =1we getg(0) =g(1 + 1) =g(1) + 1, i.e.g(1) = 1.

By the induction we can now generalize the relationg(x+ 1) = g(x) + 1 tog(x+ n) = g(x) +n for all n Z. Then it follows that g(n) = g(0 +n) =g(0) +n= n for alln Z.

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If we putx = nandy = 1

n (for0= n Z) in the relation()thengn

n

+

g

n+

1

n

=g(n)g

1

n

+g(n) +g

1

n

. Sinceg(n) =nandg

n +

1

n

=n +g

1

n

,

then we have1 + n + g1

n

=ng1

n

+ n + g1

n

, which implies thatg1

n

=

1

n .

Let nowm Z andn N. Takingx = m andy = 1n

in the relation()gives

gm

n

+ g

m +1

n

=g(m)g

1n

+ g(m) + g

1n

. This reduces tog

mn

+ g(m) +

g1

n

=g(m)g

1n

+g(m) +g

1n

, and further tog

mn

=

m

n.

Thus we have showed thatg(x) =xfor allxQ. Using the same continuityargument as in Example 9 we find thatg(x) =xfor allx R. Hencef(x) =x+1and one should now verify that this function satisfy the relation in question.

8. Takingy = 0yeldsxf(0) = xf(x2) for allx N0, i.e.f(x2) = f(0). Thismay suggest thatf(x)is a constant function. Moreover, it is clear that all constantfunctions satisfy the given equation.

Supposea, b N andf(a) < f(b). Then (a+ b)f(a) = af(a) +bf(a) 0, its safe to take logarithms on both sides. This yelds 2 ln

f(x)

=ln

f(x + y)

+ ln

f(x y). The next step is obvious: introduce a new function:g(x) = ln

f(x)

. The equation transforms to2g(x) =g(x+y) +g(x y).

In order to solve the new equation, take y=x. This gives 2g(x) =g(2x)+g(0),

i.e. g(2x) = 2g(x) g(0). Now, since2g(2x) = g(2x+ x) +g(2x x), theng(3x) = 2g(2x) g(x) = 22g(x) g(0) g(x) = 3g(x) 2g(0).

At this stage we may guess that g(nx) = ng(x) (n 1)g(0)for alln N,and we may prove this by induction.

Take now a positive rational number x = m

n, withn N. This means that

m = nx and thus, g(m) = g(nx) = ng(x) (n 1)g(0). On the other handg(m) =g(m 1) =mg(1) (m 1)g(0).

From the last two equalities we deduce that ng(x)

(n

1)g(0) = mg(1)

(m 1)g(0), i.e.ng(x) = (n m)g(0) +mg(1). Dividing both sides by n andkeeping in mind thatx =

m

n, we getg(x) = (1 x)g(0) +xg(1), which may be

written asg(x) =

g(1) g(0)x+g(0). Lettingg(1) g(0) = aand g(0) =bwe get finallyg(x) =ax+b.

The continuity ofg(x) (logarithm andf(x) are continous) allow us to extendthe result in the usual way to all x R+. Thus, f(x) = eg(x) = eax+b for anychoice of real constantsa andb.

What remains to do is to check that so obtained function satisfies the given

equation.

13. It is obvious that the identity functionf(n) = nsatisfies the given equation.We may suspect that there are no other functions than that.

First we observe that f(n) is injective. For suppose f(n) = f(m). Then ob-

viouslyf

f(n)

= f

f(m)

and consequently f

f

f(n)

= f

f

f(m)

.

Thus,f

f

f(n)

+f

f(n)

+f(n) =f

f

f(m)

+f

f(m)

+f(m), i.e.

3n= 3mandn= m.Forn = 1we getf

f

f(1)

+f

f(1)

+f(1) = 3, which can only mean

thatf(1) = 1. Hencef(2) 2,f(3) 3, and so on.Suppose now that k is the least number such that f(k) > k. Then, since

f

f(k) f(k), we would have ff(k) f(k) > k. Similarily we would

havef

f

f(k) ff(k) > k. This together would give us fff(k) +

f

f(k)

+f(k)> 3kwhich contradicts the equation.

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Hencef(k) =k for allk N

14. If the polynomialp(x) is constant,p(x) c, then, inserting it into the equationgivesc = 0 orc =1. Both of this polynomials are apparently solutions to theequation. So let us now assume thatp(x)is not constant.

Suppose x0 is a zero of p(x). Putting x0 into the equation yields p(x20) +

p(x0)p(x0+ 1) = 0, i.e. p(x20) = 0. Thus x

20 is a zero ofp(x) as well. This ar-

gument can be repeated and, by induction, one shows that x2n

0 are zeros ofp(x)for alln N. Since the polynomialp(x)has only a finite number of zeros thenx0can only equals0,1 or 1.

Letting nowx0 1into the equation yieldsp

(x0 1)2

+p(x0 1)p(x0) = 0,i.e.p

(x01)2

= 0. This means that(x01)2 is again a zero ofp(x). In the viewof the above discussion(x0

1)2 equals0,1 or

1. Hence,x0 can only equals0

or1and thenp(x) =cxn(x 1)m for somec R andm, n N. Ifc = 0, we getthe zero polynomialp(x) 0already considered. Suppose then thatc = 0

Inserting this expression into the equation givescx2n(x2 1)m + cxn(x 1)m c(x+ 1)nxm = 0, which reduces to xnm(x+ 1)mn +c = 0 for all x. Thenapparentlym = n andc = 1. Hencep(x) = xn(x 1)n for alln N.

One must now only check that these functions really satisfy the given equation.

Thus the answer isp(x) 0orp(x) 1orp(x) = xn(x 1)n for alln N.

15. Supposer, s Q such thatr < s, and letn be a positive integer. Let dividethe segment[r, s]inn equal parts byri=r+

s rn

i, fori= 0, 1, 2,...,n. Eachpart has the length |ri ri+1| = s r

n .

Then|f(r) f(s)| = n1i=0

f(ri) f(ri+1)

. Using now the triangle in-equality we get

|f(r)

f(s)

|

n1

i=0f(ri)

f(ri+1) 7

n1

i=0ri

ri+12 =7

n1i=0

s rn

2=

7(r s)2n

.

Letting now n we find out that the right hand side goes to 0 and sof(r) f(s) = 0, i.e. f(r) = f(s) for all r, s Q. Hence f(x) is a constantfunction on Q, f(x) =c for some real constantc and allx Q.

Now we can turn to the real numbers x. If then x R wa may consider a

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sequence of rational numbers {rn} Q, such that limn

rn = x. We may in fact

choose {rn} so that |x rn| 0. This is

clearly impossible, hence there is no solution in this case.

(2) Ifa >1, thenf(x) = 1

1 a2x(1 ax)

x+ 1 . Taking thenx = 12a we will get

f 1

2a

1a . Again a contradiction.

(5) If1 < a < 0, thenf(x) = 11 a2

x(1 ax)x+ 1

and this is> 0 for all

positive real numbersx.It remains to verify that this function (only for 1 < a < 0) satisfy the given

equation.

17. Takingy = x 3x+ 1

yeldsx = 3 +y1 y and the equation f(y) + f

y 3y+ 1

=

3 +y

1 y . If we instead takey =3 +x

1 x thenx=y 3y+ 1

and we get another equation:

f3 +y

1 y

+f(y) =y 3y+ 1

.

Adding now both equations together we get 2f(y)+fy 3

y+ 1

+f3 +y

1 y

=

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3 +y

1 y+y 3y+ 1

= 8y

1 y2 . On the other hand we know from the original functional

equation that f

y 3y+ 1

+f

3 +y

1

y

=y. Hence 2f(y) +y = 8y

1

y2

. Finally

f(y) = 4y1 y2

y2

= y3 + 7y2(1 y2) .

It remains to verify that this function really satisfy the given equation.

18. Supposef(m) = f(n). Then we havef(m) +f(n) =f(n) +f(n). Takingf on both sides of last equality gives, according to the relation the function fsatisfies,m+n =f

f(m) +f(n)

=f

f(n) +f(n)

=n+n. Hencem =n

and we can conclude thatf(n)is injective.

In order to find the value f(1) suppose f(1) = c > 1. Then2 = f

f(1) +f(1)

=f(2c). Thusf(2 + c) =f

f(2c) + f(1)

= 2c + 1. It is then obvious that

ccannot equals 2: puttingc = 2into the last two equalities would givef(4) = 2andf(4) = 5. Hencec >2.

Consider now the numbersf(2c) + f(1)andf(c +2)+f(c1). Applyingf tothose two numbers yieldsf

f(2c) + f(1)

= 2c + 1andf

f(c +2)+ f(c1) =

c+2+c1 = 2c+1. Since f(n) is injective then f(2c)+f(1) =f(c+2)+f(c1),which means that2 + c= 1 + 2c + f(c 1), i.e.f(c 1) = 1 c 2 must be rejected and we have c= 1.Now we claim thatf(n) = nfor alln N. We know it is true for n = 1. Sosuppose it is true for somen0 N. Thenn0+ 1 =f

f(n0) + f(1)

=f(n0+ 1).

Hence, by the induction,f(n) =n for alln N.

19. By easy induction one may extend the condition f(x+ 1) = f(x) + 1 tof(x+n) =f(x) +nfor alln N.

Consider now the positive rational number x = m

n +n2, for any m, n N.

From the second condition of the problem and using the new condition above

we get fm

n +n2

3 =

fm

n +n2

3=

fm

n

+n2

3

=

fm

n

3+

3

fm

n

2n2 + 3f

mn

n4 +n6.

On the other hand fm

n+ n2

3=f

mn

3+ 3m

n

2n2 + 3

mn

n4 + n6

=

fm

n

3+ 3m2 + 3mn3 + n6

=f

mn

3+ 3m2 + 3mn3 + n6 =

fm

n

3+

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3m2 + 3mn3 +n6.

Equating both right-hand sides gives

fm

n

3+ 3

fm

n

2n2 + 3f

mn

n4 +

n6 = fm

n 3

+ 3m2 + 3mn3 +n6, i.e. fm

n 2

n2 +fm

n n4 =m2 +mn3.

Now, it is only to discover thet the last expression can be factorised as 0 =fm

n

2n2 +f

mn

n4 m2 mn3 =

fm

n

n m

fm

n

n+m+n3

.

Since the last parenthesis is never0, thenfm

n

n m= 0, which means that

fm

n

=

m

n.

It is not difficult that the functionf(x) =x satisfies the original equation.

20. It is obvious thet the functionf(x) is not constant. The double fof the lefthand side complicates the problem considerably. To get rid of that we may first

takey = 0 (getting f

f(x)

= f(x) +f(x)f(0)) and then replacex byx+y, which results in a new equation f

f(x+ y)

= f(x+ y) +f(x+ y)f(0).

Equating the right-hand sides of this equation and the original one we get f(x+y) + f(x)f(y)xy=f(x + y) + f(x + y)f(0), orf(0)f(x + y) =f(x)f(y)xy().

Lets now try to put y = 1 into (). This will result in f(0)f(x + 1) =f(x)f(1)

x (). From the last expression we would like to eliminate the

f(x+ 1) term. In order to do that put y =1 in () (gettingf(0)f(x 1) =f(x)f(1) +x) and replace thenx by x+ 1. This gives f(0)f(x) = f(x+1)f(1) + x + 1, which multiplied byf(0)isf2(0)f(x) =f(0)f(x + 1)f(1) +f(0)(x + 1). Now we can substitute here f(0)f(x + 1) by the expression in

(): f2(0)f(x) =

f(x)f(1) xf(1) + f(0)(x + 1). Hence f2(0)f(1)f(1)

f(x) =

f(0) f(1)

x+f(0) ( ).

There are now two cases to consider: the coefficient on the left hand side equals

0 or not.

Supposef2(0) f(1)f(1) = 0. Then putting x = 0 in ( ) results inf(0) = 0. Hence,f(1)f(1) = 0. At the same time the equalityf(0) = 0turn ()intof(x)f(y) = xy. Takingx = 1and y =1we getf(1)f(1) =1, whichcontradicts the previous result.

Suppose finally thatf2(0) f(1)f(1)= 0. Then the expression ( ) im-plies that f(x) is a polynomial of degree one, f(x) = ax+ b. Substituting thispolynomial into the original equation we geta

a(x+y) +b

+b = a(x+y) +

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b+ (ax+b)(ay+b) xy. Since this is valid for allx, y R the by taking somevalues forx andy it is easy now to show that a = 1 andb = 0. Thus, the onlysolution to the equation isf(x) =x.

21. Since the conditionf(m+n) f(m) f(n) = 0or 1 is not easy to handlewe may try to replace it with a (weaker) condition f(m+n) f(m) +f(n). Solets see how far do we get.

We begin with finding f(1): 0 = f(2) = f(1 + 1) f(1) +f(1) = 2f(1).Sincef(1) 0the we havef(1) = 0.

Now,f(3) = f(2 + 1) = f(2) +f(1) +a = a, wherea equals 0 or 1. Sincewe know thatf(3)> 0 then, of course,f(3) = 1

Next we may note that f(2 3) = f(3 + 3) f(3) +f(3) = 2,f(3 3) =f(2

3 + 3 )

f(2

3) + f(3)

3, and generally, by induction, thatf(3

n)

n.

Moreover, if we for somek get f(3k)> k, then the same argument shows thatf(3m) > m for all m > k. But we know that f(9999) = f(3 3333) = 3333,hencef(3n) =n for alln upp to at least 3333.

Now,1982 = f(3 1982) = f(2 1982 + 1982) f(2 1982) +f(1982) =f(1982 + 1982) +f(1982) 3f(1982), implying that f(1982) 660. On theother hand,f(1982) =f(1980+2) f(1980) + f(2) =f(3 660) = 660. Thus,f(1982) = 660.

22. It is obvious that f(1) = 1. Ifm is an odd integer then (m, 2) = 1 and wehavef(2m) =f(2)f(m) = 2f(m). Hence, ifmis an odd integer andf(m) =m,thenf(2m) = 2m.

Lets try to find the value off(3). There are many ways of doing this, forexample through the following, rather artificial, reasoning (remember thatf(x)isstrictly increasing):

2f(7) = f(2)f(7)< f(3)f(7) = f(21) < f(22) = f(2)f(11) = 2f(12) n0and morover, sincef(x)is strictly increa-sing,f(n)> nfor alln n0 ().

Let now then consider two cases:

(1) Ifn0 is odd then (2, n0 2) = 1 and so2(n0 2) = f(2)f(n0 2) =

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f

2(n0 2)

. However, for n0 > 3, 2(n0 2) n0 and so, according to (),f

2(n0 2)

>2(n0 2), giving2(n0 2)> 2(n0 2), thus a contradiction.(2) Ifn0 is even then(2, n0 1) = 1 and so2(n0 1) = f(2)f(n0 1) =

f2(n01). Again, forn0>3,2(n0

1)

n0and sof2(n0

1) >2(n01),

giving2(n0 1)> 2(n0 1), a contradiction.In conclusion, suchn0doesnt exist and hencef(n) =n for alln N.

23. Lets try some values ofx andy. Takingx = y = 1 yields f(1) +f(1) =f(3) +f(1), but sincef(1) =f(1)thenf(3) =f(1).

Taking nowx = 1 andy = 2 givesf(1) +f(2) = f(5) +f(2), but if wetakex = 2andy =1thenf(2) +f(1) = f(2) + f(3). From the two lastexpression we find thatf(5) =f(3).

Since thenf(1) = f(3) = f(5)we may suspect that f(n)has the same valuefor all odd integers.

This is in fact correct and in order to prove it just take first x = 1and y = m(giving f(1) +f(m) = f(1 + 2m) +f(m)) and thenx = mand y =1(sowe getf(m) + f(1) =f(m) + f(1 + 2m). From the two last expression wefind thatf(2m 1) =f(2m+ 1)for allm Z.

Another consequence of the equationf(1) +f(m) =f(1 + 2m) +f(m)) isnow thatf(m) = f(m). Since thenf(x)ie an even function, it is suficcient tofind the expression forf(x)for non-negative integersx.

Let nowx = n andy =(2k+ 1). Then our equation implies f(n) + f (2k+ 1)

=f

n(1 + 4k)+f (2k+ 1)(1 2n), which, after cancellingf(m) for odd m, means that f(n) = f

n(1 + 4k) = f(1 + 4k)n. If weinstead takex= (2k+ 1)andy =n then f (2k+ 1)+ f(n) =f (2k+1)(1+2n)

+ f

n(4k+ 3)

, i.e. f(n) =f

(4k+ 3)n

. Thusf(n) =f(mn)forany odd inegerm.

Every positive integerncan be written in formn= 2amfor some non-negativeinteger a and an odd integer m. Hence f(n) = f2am = f2a. Thus anyfunction with of the kind we are looking for is determined by the values f(0),f(1),f(2),f

22

,f

23

,f

24

and so on, which may all be chosen arbitrary. All

other values are given byf(n) =f

2a

as above. For negative integersk we havethatf(k) =f(k).

Finally lets check that such functions satisfy the equations. Clearly f(1) =f(1). Ifx= 0ory= 0then the equation becomes an identity. So suppose thatx=2amand y = 2bnfor some non-negative integersa and b and oddm,n. Then the

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left-hand side of the equation becomes f(2am) +f(2bn) =f(2a) +f(2b), whilethe right-hand side becomes f

2am(1+2y)

+ f

2bn(1 2x) =f(2a) + f(2b),since bothm(1 + 2y)andn(1 2x)are odd.

Thus the solution is complete.

24. Since f : N N, it ishould be clear that f(1) > 1. Moreover, f(2) >f(f(1)) 1, which implies that f(2) 2. The same argument cannot howeverbe extended for showing thatf(3) 3.

Nevertheless it is possible to prove slightly stronger statement, from which the

inequalityf(n) nfollows immediately.Statement: Ifm nthenf(m) n.This statement is obviously true forn = 1sincef(m) 1by the definition of

f(m). So let us assume that the statement is true for some n0

1, i.e. assume thatIfm n0thenf(m) n0.

Now, suppose thatm n0+1. hence,m1 n0and then, by the assumption,f(m 1) n0. By the assumption again,f

f(m 1) n0. Using the property

off(m) in the statment of the problem, we know that f(m) > f

f(m 1).Hence f(m) > f

f(m 1) n0. This means that f(m) n0+ 1 and, by

induction, the statment is true for alln N.As a special case, we have f(n) n for all n N. From this it follows that

f(n+ 1)> f(f(n))

f(n), proving that the functionf(n)is strictly increasing.

Finally, suppose thatf(n) =n for somen N. Then, of course,f(n)> nandwe getf(n)n+ 1. This implies thatf(n+ 1)> f(f(n))f(n+ 1), whichis impossible.

Therefore,f(n) =n for alln N.

25. As in the solution of Example 19, we discover that the constant function

f0(x) 1 is a solution to the equation, and then that for non-constant solutionf(x)we havef(xy) =f(x)f(y)for allx, y R+.

Now we find thatf(a)f(x+y)

= f

ax+y

=f

ax

ay

=(by the previous equa-lity)= f

ax

f

ay

= f(a)f(x)f(a)f(y) = f(a)f(x)+f(y), from which follows thatf(x+y) =f(x) +f(y)for allx, y R+.

By the same methods as in Example 9 one can show now that f(x) = xf(1)for allx Q+. At the same time, settingx = y = 1 into the equationf(xy) =f(x)f(y), we getf(1) = 1, and thusf(x) =x for allx Q+. What then remainsis to extend this formula to allx R+.

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Suppose that for some x > 0 we have f(x) < x (in the case f(x) > x theargument is similar). Lets pick up a rational number a such thatf(x) < a < x.Then we will havef(x) =f

a + (x a) =f(a) + f(x a)> f(a) =a, which

contradicts the choice ofa. Hencef(x)

1and f(x) = xare the only solutions

to the problem.

26. By letting x = y = 0 and u = v, we get 4f(0)f(u) = 2f(0). So eitherf(u) = 1/2 for all u R, or f(0) = 0. The constant function f(u) = 1/2 iscertainly a solution. Hence assume thatf(0) = 0.

Puttingy=v = 0we getf(x)f(u) =f(xu) (). In particular, takingx= u =1, we havef(1)2 = f(1). Hencef(1) = 0orf(1) = 1. Suppose thatf(1) = 0.By takingx = y = 1andv = 0, we get0 = 2f(u). Thusf(x) = 0for allu R.That is certainly a solution as well. We can thus assume that f(1) = 1.

Settingx = 0and u = v = 1, we get2f(y) = f(y) +f(y), which reducestof(y) = f(y). This means that f(x)is an even function and so we need onlyconsiderf(x)for positivex.

Next we show that f(r) = r2 for all r Q. The first step is to show thatf(n) = n2 for alln N. This is done by the induction on n. It is obviously truefor n = 0 and 1. Suppose it is true for n 1 and n. Then letting x = n andy=u = v = 1into the equation, we get2f(n) + 2 =f(n 1) + f(n + 1). Hencef(n+ 1) = 2n2 + 2

(n

1)2 = (n+ 1)2. Hence the statment is true forn+ 1.

Now the relation () implies thatf(n)fm

n

= f(m), sof

mn

= m

2

n2 for all

m, n N. Hence we have established thatf(r) = r2 for allr Q+. By the factthatf(x)is even,f(r) =r2 for allr Q.

Now it is natural to suspect that f(x) = x2 for allx R, so this is what weshould try to prove in the final step. Since we dont have the condition thatf(x)iscontinous, we cannot make use of the standard procedure for those cases.

From the relation () above, we havef(x2) = f(x)2 0, sof(x2)is alwaysnon-negative. Hence f(x) 0 for positive x and, again by the fact that f(x) iseven,f(x) 0for allx R.

Putting now u = y and v = x, we get

f(x) +f(y)2

= f(x2 +y2), sof

x2 + y2

=f(x)2 + 2f(x)f(y) + f(y)2 f(x)2 =f(x2). For anyuandvsuchthatu > v >0, we may putv =x2 andu =x2 +y2, and hencef(u) f(v). Inother words,f(x)is an increasing function.

Thus for any realxwe may take a sequence of rationalsrn, all less thanx, thatconverge tox and another sequence of rationalssn, all greater thanx, which also

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converge tox. Then we get r2n = f(rn)f(x)f(sn) = s2n for allx R andhencef(x) =x2.

The final answer is then: there are three possible functions solutions, namely

f(x) = 0for allx

R,f(x) =

1

2

for allx

R orf(x) =x2.

ADDITIONAL PROBLEMS

Here follows some more problems, this time without solutions offered. Instead,

after the problems there are some hints and answers.

Problems.

27. (Poland, 1992) Find all functionsf : R R such thatf(x + y)f(xy) =f(x)f(y)for allx, y R.

28. Find all functions that satisfy the equationf(1

x) +xf(x

1) =

1

x for all

realx = 0,x = 1andx = 1.29. Find all continous functionsf : R R which satisfy the equationf(x+y) =f(x) +f(y) +xy for allx, y R.

30. Find all functionsf : R R satsifyingxf(y) +yf(x) = (x+y)f(x)f(y)for allx, y R.

31. Find all injective (one-to-one) functionsf : R

R such thatff(x) +y =

f(x+y) + 1 for allx, y R.32. Find all functions f : Z Z satisfying following conditions f(1) = 1 andf(x+y)

f(x) f(y) =f(x y)f(x) +f(y) for allx, y Z.

33. Find all polynomialsp(x)satisfying the equationp(x2 2x) = p(x 2)2for allx R.

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34. Find all continous functionsf : R R such thatf(1) = 1andf

x2 +y2 =f(x) +f(y)for allx, y R.

35. Find all functions defined forx >0, such thatxf(y) + yf(x) =f(xy)for all

x, y R+.36. Find all continous solutionsf : R R to the equationf(x + y)f(xy) =f(x)for allx, y R.

37. Find all functionsf : R R, continous inx = 0which satisfy the equationf(x+y) = f(x) + f(y) + xy(x+y) for all x, y R. (Compare with problem25.)

38. Find all functionsf : R R which solve the equationf(x + y) + f(xy) =2f(x)cos y for allx, y R.

39. Supposef : N N is a strictly increasin function such that f(f(n)) = 3nfor alln N. Find all possible values off(1977).

40. (AMM, Problem E2176) Find all functionsf : Q Q such thatf(2) = 2andf

x+y

x

y =

f(x) +f(y)

f(x)

f(y)

for all rationalx =y.

41. (Austria-Poland, 1997) Show that there is no functionf : Z Z such thatf

x+f(y)

=f(x) y for allx, y Z.

42. (Ukraine, 1997) Find all functionsf : Q+ Q+ such that f(x+ 1) =f(x) + 1 and f(x2) =

f(x)

2for allx Q+.

43. (IMO short-list, 1999) Suppose that the functionf : R R satisfies twoconditions:

|f(x)

| 1 for all x

R and fx+ 13

42+f(x) =fx+ 1

6+fx+ 1

7

for allx R. Prove thatf(x)is periodic.

44. (IMO, 1981) The functionf(x, y)satisfies:f(0, y) =y+ 1, f(x + 1, 0) =f(x, 1)andf(x + 1, y+ 1) =f(x, f(x + 1, y))

for all non-negative integersx, y. Findf(4, 1981).

45. (IMO, 2004) Find all polynomialsP(x)with real coefficients which satisfy

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the equality

P(a b) +P(b c) +P(c a) = 2P(a+b+c)for all real numbersa, b, csuch thatab+bc+ca= 0.

46. (IMO, 1994) LetSbe the set of all real numbers greater than 1. Find allfunctionsf: S Ssuch thatfx+f(y) +xf(y) =y+f(x) +yf(x)for allx, y S, and f(x)

x is strictly increasing on each of the intervals 1< x


43/46

Answer:f(x) 0.

28. Hint: Create a new equation.

Answer:f(x) = 3 xx(1 x2)

.

29. Hint: One way of solving is the standard procedure, findingf(n)for integersn, extend it to Q and then to R.

Another way is to guess that f(x) =1

2x2 is one solution. Are there more?

Answer:f(x) = 12x2 +cxfor any real constantc.

30. Hint: Try some values.

Answer:f(x) 0orf(x) =

1 if x = 0c if x= 0

for any real constantc.

31. Hint: Letx andy change places.

Answer:f(x) =x+ 1.

32. Hint: Show thatf(n)is odd. Take thenx = 2andy = 1and consider somecases.

Answer:f(n) = 0 if n= 2k1 if n= 4k+ 11 if n= 4k+ 3 , or f(n) =

0 if n= 3k1 if n= 3k+ 11 if n= 3k+ 2 ,

fork Z, orf(n) =n for alln Z.

33. Hint: Solve first the functional equationq(x2) =

q(x)2

, where q(x) is apolynomial.

Answer:p(x) 0or p(x) = (x+ 1)n for eachn N.

34. Hint: Find the expression forf(x)first forxin N, then in Q.

Answer:f(x) =x2.

35. Hint: Transform to a Cauchy-type equation.

Answer:f(x) =cx ln xfor any real constantc.

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36. Hint: Standard procedure.

Answer:f(x) =cx for any real constantc.

37. Hint: One way of solving is the standard procedure, findingf(n)for integers

n, extend it to Q and then to R. But first you will have to show that the continutyinx= 0will imply thatf(x)is continous for allx R.

Another way is to discover that 3xy(x+y) is a part of the expression for (x+y)3

and thus guess thatf(x) =1

3x3 is one solution. Are there more?

Answer:f(x) = 13x3 +cxfor any real constantc.

38. Hint: Start with some values forxandy.

Answer:f(x) =a cos x+b sin xfor any choice ofa, b R.39. Hint: Prove that f(3k) = 3f(k). Show then that for 3m n < 23mthe function is f(n) = n+ 3m, while for 2 3m n < 3m+1 one must havef(n) = 3n 3m+1.

Answer:f(1997) = 3804.

40. Hint: Findf(0)andf(1). Show then that fm

n = f(m)

f(n) for all m, n Z,

n = 0.Answer:f(x) =x.

41. Hint: Show that g(x) = f

f(x)

is injective (in fact linear) and then show

thatf(x)satisfy the first of Cauchys equations. Find then the contradiction to theexistens of the solution.

42. Hint: Countfm

n +n

2

in two different ways.Answer:f(x) =x.

43. Hint: Show that the functiong(x) =f(x +1

6)f(x)is periodic and then that

the functionh(x) =f(x+ 1) f(x)is periodic.Answer: The sortest period forf(x)is 1.

44. Hint: Calculatef(1, n),f(2, n),f(3, n)and find the pattern.

44


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Answer: f(4, 1981) = 222...2

3, a tower of 1984 2s less 3. In general, f(4, n) =22

2...2

3, a tower of(n+ 3) 2s less 3.

45. Hint: For avery real number t the triple (a,b,c) = (6t, 3t, 2t) satisfy theconditionab+bc+ca= 0. What implication does it have on the equation?Answer:P(x) =x4 +x2 for any choice of real numbers and.

46. Hint: Start by takingy =x. Find out thatx + f(x) + xf(x)is a fixed point off(x)for eachx S. How many fixed points canf(x)has at most?

Answer:f(x) = xx+ 1

.

47. Hint: Prove that f(0) = 0. Show that f(f(y)) = y for all real y. Showthereafter thatf(x+y) =f(x) +f(y).

Answer:f(x) =x.

48. Hint: Start withy =x and then changex to f(x). Assume thatf

f(x)

> xand dont forget to use the fact that f(x)is strictly decreasing.

49. Hint: Think of the numbers in base 2, i.e. let n2 be the binary representation

ofn. Prove thereafter (using the induction) that the function f(n2) returns thenumberm2which has the same digits asn2but in the opposite order. The problemreduces then to finding the number of all integers 1988 with the symmetricbinary representation. Find that the number of symmetrical binary numbers with

kdigits is2(n1)/2. How many (binary) digits do we need in order to not exceed1988?

Answer:92.

50. Hint: Show that f(x) = 0 for all x

2. How should f(x) look like for0 x


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the smalest value off(1998)we may assumef(1) = 1.Show then that ifp is a prime number thatf(p)is a prime number as well and

f(f(p)) =p. Show finaly thatf(n)can be defined arbitrary on primes as long asthe conditionsf(p) =q (wherep andqare prime) andf(q) =p are satisfied.

Answer:2 2 2 3 5 = 120.

52. Hint: Show that f(xy) = f(x)f(y) and f

f(x)

= 1

x for all x, y Q+.

Forx Q+ construct f(x) based on the prime faktorization of numerator anddenumerator: x = pn11 p

n22 ...p

nkk , where pi are prime numbers and ni Z, since

we havef(x) =

f(p1)n1f(p2)n2...f(pk)nk (). Thus it is enough to define

a suitable function on the set of all prime numbers {p1, p2, p3,...} and, of course,addf(1) = 1.

Answer: One possible construction isf(pj) =

pj+1 if j is odd,

1pj+1

if j is even ,f(1) = 1

and extend it to whole Q+ using ().

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Vaderlind P. the Quest for Functions. Functional Equations for the Beginners (2005)(46s)_MSch

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