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Eisenstein’s Criterion Applied to mth OrderBernoulli Polynomials of Degree m
by
Michael Peretzian Williams
Bachelor of Science, MathematicsUniversity of South Carolina, Columbia, SC December 1998
Submitted in Partial Fulfillment of the Requirements
for the Degree of Master of Science in the
Department of Mathematics
University of South Carolina
August 2001
Department of MathematicsDirector of Thesis
Department of MathematicsSecond Reader
Dean of The Graduate School
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Abstract
The generalized Bernoulli polynomials B(`)m (x) are defined by(
t
et − 1
)`
etx =∞∑
m=0
B(`)m (x)
tm
m!.
The polynomial B(`)m (x) is of degree m and ` is referred to as its order. This thesis is
concerned with the polynomials B(m)m (x) (i.e. the case when ` = m). We show that a
positive proportion of these are Eisenstein. More precisely, we show that asymptoti-
cally for at least 1/5 of these polynomials B(m)m (x), there is a prime p (depending on
m) such that p does not divide the numerator or denominator of the leading coeffi-
cient of B(m)m (x), p divides the numerator but not the denominator of all of the other
coefficients, and p2 does not divide the numerator of the constant term. (Here we are
viewing the coefficients as reduced fractions.) Some background material is discussed
in the course of establishing this result.
ii
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Contents
Abstract ii
Chapter 1 Recurrence Relations for Bernoulli Numbers and Polynomials 1
Chapter 2 An Explicit Formula and the von Staudt-Claussen Theorem 7
Chapter 3 Higher Order Bernoulli Polynomials and Eisenstein’s Criterion 14
iii
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Chapter 1
Recurrence Relations for Bernoulli Numbers
and Polynomials
Definition 1. The Bernoulli polynomials Bm(y) are given by
tety
et − 1=
∞∑m=0
Bm(y)tm
m!.
Observe that Definition 1 implies that Bm(y) is a polynomial of degree m with
rational coefficients. Integrating the left-hand side above with respect to y and eval-
uating from x to x + 1 yields∫ x+1
x
tety
et − 1dy =
t
et − 1
∫ x+1
x
etydy
=t
et − 1
[1
tety
]x+1
x
=etx(et − 1)
et − 1
= etx
=∞∑
m=0
xm tm
m!.
And now by taking the same integral of the right-hand side and comparing it to the
above we get∞∑
m=0
(∫ x+1
x
Bm(y)dy
)tm
m!=
∞∑m=0
xm tm
m!.
Comparing like terms we get the following [HE]:
(1)
∫ x+1
x
Bm(y)dy = xm.
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Now we show that the polynomials Bm(y) are uniquely determined by (1) for all
m. Let
f(x) =n∑
j=0
ajxj where an 6= 0
be such that ∫ x+1
x
f(t)dt = xm.
Taking the derivative of both sides we obtain
f(x + 1)− f(x) = mxm−1.(2)
We deduce that (1) uniquely determines the polynomials Bm(y) from (2) and the
next lemma.
Lemma 1. Let f(x) and g(x) be polynomials in C[x] such that
f(x + 1)− f(x) = g(x + 1)− g(x).(3)
Then f(x) = g(x) + c for some constant c.
Before proving the lemma we note that it implies that f(x) in (2) is unique up to
a constant. Given (1), Bm(y) = f(y) + c for some c. As∫ x+1
x(Bm(y)− f(y)) dy =
xm−xm = 0, we deduce c = 0. Thus, (1) uniquely determines the polynomial Bm(y).
We now give two proofs of Lemma 1.
Proof 1. Observe that
f(x + 1)− f(x) =n∑
j=0
(aj(x + 1)j − ajx
j)
(4)
=n∑
j=1
aj
j−1∑k−0
(j
k
)xk
=n−1∑k=0
(n∑
j=k+1
aj
(j
k
))xk.
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As the coefficient of xn−1 is nan 6= 0, we deduce that the degree of f(x + 1)− f(x) is
one less than the degree of f(x). Hence, (3) implies that the degree of g(x) is n. Let
h(x) =n−1∑j=0
bjxj,
and suppose f(x+1)−f(x) = h(x). To establish the lemma it suffices to show that the
aj’s with j > 0 are uniquely determined by the bj’s. From (4) and f(x + 1)− f(x) =
h(x), we obtain
a1 + a2 + a3 + . . . + an−2 + an−1 + an = b1
2a2 + 3a3 + . . . + (n− 2)an−2 + (n− 1)an−1 + nan = b2(3
2
)a3 + . . . +
(n− 2
2
)an−2 +
(n− 1
2
)an−1 +
(n
2
)an = b3
...(n− 2
n− 3
)an−2 +
(n− 1
n− 3
)an−1 +
(n
n− 3
)an = bn−2(
n− 1
n− 2
)an−1 +
(n
n− 2
)an = bn−1(
n
n− 1
)an = bn.
Note that the matrix of coefficients for this system of equations is an upper tri-
angular matrix, with a non-zero determinant. Indeed, the determinate is
n∏j=1
(j
j − 1
)=
n∏j=0
(j
1
)= n! 6= 0.
Hence, the aj’s are determined uniquely from the above system of equations. There-
fore, the lemma follows. �
Proof 2. Let h(x) = f(x)− g(x), and set c = h(0). We wish to show h(x) ≡ c.
From (3), h(x+1) = h(x). We consider w(x) = h(x)−c, and note that w(x+1) = w(x)
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and w(0) = 0. It follows that
0 = w(0) = w(1) = w(2) = · · ·.
In other words, w(n) = 0 for every integer n ≥ 0. Since f(x) and g(x) are polynomials,
w(x) is a polynomial with infinitely many zeroes. Hence, w(x) ≡ 0, and we obtain
h(x) ≡ c as desired. �
From (1) the following useful results emerge by taking the derivative [HE]:
Bm(x + 1)−Bm(x) = mxm−1(5)
Bm(x + 1)−Bm(x)
m= xm−1(6)
Bm(x + 1)−Bm(x) = m
∫ x+1
x
Bm−1(t)dt.(7)
Observe that if G(x) is an anti-derivative of mBm−1(x), then
G(x + 1)−G(x) = m
∫ x+1
x
Bm−1(t)dt.
From (7) and Lemma 1, we obtain that Bm(x) is an anti-derivative of mBm−1(x).
Definition 2. The number Bj = Bj(0) is the jth Bernoulli number.
From (1),
∫ x+1
x
B0(t)dt = x0 = 1.
Since B0(x) is a constant polynomial, we deduce B0(x) = 1 so that B0(0) = 1 and,
by definition, B0 = 1.
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We use that Bm(x) is an anti-derivative of mBm−1(x), the definition above and
B0 = 1 to obtain the following:
B1(x) =
∫B0(x)dx = x + B1
B2(x) = 2
∫B1(x)dx = x2 + 2B1x + B2
B3(x) = 3
∫B2(x)dx = x3 + 3B1x
2 + 3B2x + B3
...
An induction argument gives that in general
(8) Bm+1(x) = (m + 1)
∫Bm(x)dx =
m+1∑k=0
(m + 1
k
)Bkx
m+1−k.
Observe that (8) implies
Bm+1(x) = (x + B)m+1
where Bj in the expansion of (x + B)m+1 is interpreted as the value of Bj (and not a
power). We will generalize this later. We now set out to find a recursive formula for
the Bernoulli numbers. Consider (1) and substitute 0 for x. For m > 0 we have
0m = 0 = (m + 1)
∫ 1
0
Bm(t)dt
= Bm+1(1)−Bm+1(0) (by (7))
=m+1∑k=0
(m + 1
k
)Bk −Bm+1 (by (8) and the definition of Bm+1)
=
(m−1∑k=0
(m + 1
k
)Bk + Bm+1 + (m + 1)Bm
)−Bm+1
=m−1∑k=0
(m + 1
k
)Bk + (m + 1)Bm.
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Thus,
Bm = − 1
m + 1
m−1∑k=0
(m + 1
k
)Bk.(9)
Using that B0 = 1 and the recurrence relation above we deduce that the first
few Bernoulli numbers are: B1 = −12
, B2 = 16, B3 = 0, B4 = −1
30, B5 = 0, B6 = 1
42,
B7 = 0, B8 = 142
, B9 = 0, B10 = 566
, B11 = 0, B12 = −6912730
, B13 = 0, B14 = 76, B15 = 0,
B16 = −3617510
, B17 = 0, B18 = 43,867798
. Note that B2m+1 appears to be 0 for m > 0. In
the next chapter we will show that this is indeed the case.
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Chapter 2
An Explicit Formula for the Bernoulli
Numbers and the von Staudt-Claussen Theorem
Theorem 1. The Bernoulli number Bn is given explicitly by the following:
A) B1 = −12
and B0 = 1.
B) Bn = 0 if n = 2m + 1 for m > 0.
C) Bn = (−1)m−1 2(2m)!)(2π)2m ζ(2m) if n = 2m > 0.
Proof. Part A comes from the previous chapter. Next we prove Part B. We will
show −B2m+1(1 − y) = B2m+1(y). Observe that Bm(0) = Bm(1) for m > 1 by (5).
Taking y = 0, we can then deduce for m > 0 that
B2m+1 = B2m+1(0) = −B2m+1(1) = −B2m+1(0) = −B2m+1
so that B2m+1 = 0.
To establish Part B, it remains to prove that B2m+1 (1 − y) = −B2m+1(y).
Recalling Definition 1 and plugging in 1− y for y yields
∞∑m=0
Bm(1− y)tm
m!=
tet(1−y)
et − 1
=tete(−t)y
et − 1
=te(−t)y
1− e(−t)
=(−t)e(−t)y
e(−t) − 1
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=∞∑
m=0
Bm(y)(−t)m
m!
=∞∑
m=0
(−1)mBm(y)tm
m!.
Now comparing like terms we get
Bm(1− y) = (−1)mBm(y).
Hence
−B2m+1(1− y) = B2m+1(y).
as desired.
Now we prove Part C using a proof directly from [BS]. From Definition 1 and
Bm = Bm(0), we deduce
t
et − 1=
∞∑m=1
Bmtm
m!.
We use a formula for the cotangent, namely
cot z =1
z+
∞∑n=1
2z
z2 − (πn)2.
On the other hand,
cot z = ieiz + e−iz
eiz − e−iz
= i +2i
e2iz − 1.
Therefore,
i +2i
e2iz − 1=
1
z+
∞∑n=1
2z
z2 − (πn)2.
We now set z = t2i
to obtain
i +2i
et − 1=
2i
t+
∞∑n=1
4it
t2 + (2πn)2.
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Multiplying both sides by t2i
and doing some algebra we get
t
et − 1= 1− t
2+ 2
∞∑n=1
t2
t2 + (2πn)2
= 1− t
2+ 2
∞∑n=1
( t2πn
)2
( t2πn
)2 + 1.
Letting t2πn
= x, the summand becomes x2
x2+1. Since
x2
x2 + 1=
∞∑m=1
(−1)m−1x2m,
we obtain
t
et − 1= 1− t
2+ 2
∞∑n=1
∞∑m=1
(−1)m−1
(t
2πn
)2m
.
Now switching the order of summation we obtain
t
et − 1= 1− t
2+ 2
∞∑m=1
∞∑n=1
(−1)m−1
(t
2π
)2m(1
n
)2m
= 1− t
2+
∞∑m=1
(−1)m−1 2ζ(2m)
(2π)2mt2m
= 1− t
2+
∞∑m=1
(−1)m−1 2ζ(2m)(2m)!
(2π)2m(2m)!t2m.
From Definition 1 (with y = 0), we now deduce
t
et − 1=
∞∑m=1
Bmtm
m!
= 1− t
2+
∞∑m=1
B2mt2m
(2m)!
= 1− t
2+
∞∑m=1
(−1)m−1 2ζ(2m)(2m)!
(2π)2m× t2m
(2m)!.
Hence, B0 = 1, B1 = −12, B2m+1 = 0 for m ≥ 1, and
B2m = (−1)m−1 2ζ(2m)(2m)!
(2π)2m.
This completes the proof. �
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Theorem 2 (The Von Staudt-Clausen Theorem). Let Bm be the mth Bernoulli
number as defined above where m is an integer > 1, and let Bm = Nm
Dmwhere Nm and
Dm are relatively prime integers with Dm > 0.
A) If p is an odd prime such that (p− 1)|m, then p||Dm and pBm ≡ −1 (mod p).
B) If p is an odd prime such that (p− 1) - m, then pBm ≡ 0 (mod p).
For this proof, we now define Sm(n) = 0m + 1m + 2m + . . . + (n− 1)m. From (1),
we obtain the following:
0m =
∫ 1
0
Bm(t)dt
1m =
∫ 2
1
Bm(t)dt
2m =
∫ 3
2
Bm(t)dt
...
(n− 1)m =
∫ n
n−1
Bm(t)dt
Hence,
Sm(n) =
∫ n
0
Bm(t)dt.
Now by (8), we see that
(10) Sm(n) =1
m + 1
m∑k=0
(m + 1
k
)Bkn
m+1−k.
The following lemma will be useful in the proof of Part A.
Lemma 2. If r is the number such that pr||n!, then r =[
np
]+[
np2
]+[
np3
]+ · · · .
We omit the proof of Lemma 2. The result is well-known and not difficult to
establish.
Proof. We prove the results of Part A and B of the theorem using induction
and following closely the proof in [BS]. If m = 2, then 1 and 2 are our only positive
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integers that divide 2, and 1 + 1 = 2 is a prime and 2 + 1 = 3 is a prime. Let p1 = 2
and p2 = 3. Recall that B2 = 16
so D2 = 6 = p1p2 and consequently p1||D2 and
p2||D2. Also,
p1B2 = 2× 1
6=
1
3≡ 1 ≡ −1 (mod 2)
p2B2 = 3× 1
6=
1
2≡ 2 ≡ −1 (mod 3).
If p is an odd prime and (p− 1) - 2, then p > 3 and pB2 ≡ 0 (mod p). Thus, Part A
and B are true if m = 2.
Consider now m > 2 and assume pBj can be expressed as a rational number with
denominator not divisible by p for j < m. We show pBm is also such a rational
number and that
pBm ≡
−1 (mod p) if (p− 1)|m
0 (mod p) if (p− 1) - m.
Theorem 2 will then follow. We know from (10) that
Sm(p) =1
m + 1
m∑k=0
(m + 1
k
)Bkp
m+1−k
= pBm +1
m + 1
m−1∑k=0
(m + 1
k
)pm−k(pBk)
so that
pBm = Sm(p)− 1
m + 1
m−1∑k=0
(m + 1
k
)pm−k(pBk)(11)
= Sm(p)−m−1∑k=0
m!
k!(m− k + 1)!pm−k(pBk)
= Sm(p)−m−1∑k=0
m(m− 1) . . . (k + 1)
(m− k + 1)!pm−k(pBk).
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Let k ∈ {0, 1, . . . ,m − 1}, and let β = β(k) be the largest non-negative integer
such that pβ|(m− k + 1)!. By Lemma 1,
β =
[m− k + 1
p
]+
[m− k + 1
p2
]+
[m− k + 1
p3
]+ . . .
<m− k + 1
p+
m− k + 1
p2+
m− k + 1
p3+ . . .
=m− k + 1
p− 1
≤ m− k + 1
2
≤ m− k.
By the induction hypothesis, we also have pBk is a rational number with denominator
not divisible by p. We deduce that each summand appearing in the last sum in (11)
is a rational number which, when reduced, has numerator divisible by p. It follows
that the entire sum is also such a rational number. Since Sm(p) ∈ Z, (11) implies
pBm is a rational number with denominator not divisible by p. Also, from (11) and
the definition of Sm(p), we obtain
pBm ≡ 1m + 2m + 3m + · · ·+ (p− 1)m (mod p).
If m = (p− 1)b for some integer b > 0, then
pBm ≡ 1b(p−1) + 2b(p−1) + 3b(p−1) + . . . + (p− 1)b(p−1) (mod p)
≡ 1b + 1b + 1b + . . . + 1b (mod p)
≡ p− 1 ≡ −1 (mod p).
If (p− 1) - m, then let g be a primitive root modulo p. We obtain
pBm ≡p−1∑j=1
jm ≡p−2∑j=0
gjm (mod p)
≡ g0·m + gm + g2m + . . . + g(p−3)m + g(p−2)m
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≡ g(p−1)m − 1
gm − 1
≡ 1m − 1
gm − 1
≡ 0 (mod p).
This completes the proof. �
Comment: In our argument above, we showed that β < m− k. For this part of our
argument, we used that p is an odd prime. We note here what happens in the case
that p = 2. Instead of β < m − k we obtain β < m − k + 1 so that β ≤ m − k. If
m > 1 is odd and p = 2, then (p− 1)|m but Nm = 0 and Dm = 1 so that we cannot
deduce as in Part A of Theorem 2 that p||Dm or that pBm ≡ −1 (mod p). However,
the situation is different if m > 1 is even and p = 2. Here, (p − 1)|m and p|m. In
each term of the final sum in (11), the numerator of the fraction is divisible by m
and, hence, p. It follows from β ≤ m − k that each term in this sum is as before a
rational number which, when reduced, has numerator divisible by p. The remainder
of the argument showing pBm ≡ −1 (mod p) follows as before. Thus, we deduce that
Part A still holds in the case p = 2 provided we add the condition that m is even.
Part B does not apply as p = 2 implies that (p− 1)|m for every integer m.
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Chapter 3
Higher Order Bernoulli Polynomials and
Eisenstein’s Criterion
In this chapter, we consider Bernoulli polynomials of higher order.
Definition 3. The generalized Bernoulli polynomials B(`)m (x) are defined by(
t
et − 1
)`
etx =∞∑
m=1
B(`)m (x)
tm
m!.
The goal of this chapter is to establish the following.
Theorem 3 (Main Theorem). Asymptotically more than one-fifth of the polyno-
mials B(`)m (x) are irreducible (and in fact Eisenstein). More precisely,
lim infu→∞
|{m ≤ u : B(m)m (x) Eisenstein}|
u>
1
5.
The definition of Eisenstein is given momentarily. We first demonstrate how we
can calculate B(`)m (x) from Definition 3. These can be calculated using the Taylor
expansion, i.e.
(B0 + B1t + B2t2
2!+ · · · )`(1 + tx +
(tx)2
2!+ · · · ).
For example, if we want to find B(3)2 (x), we expand
(B0 + B1t + B2t2
2!+ · · · )3(1 + tx +
(tx)2
2!+ · · · )
and look at the coefficient of t2, which is, x2 − 3x + 2. For a second example, let’s
find B(4)5 (x). Now we expand
(B0 + B1t + B2t2
2!+ · · · )4(1 + tx +
(tx)2
2!+ · · · )
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and look at the coefficient of t5, which is
x5 − 10x4 +110
3x3 − 60x2 +
251
6x− 9.
Now let’s find B(7)7 (x) in the same way. We look at
(B0 + B1t + B2t2
2!+ · · · )7(1 + tx +
(tx)2
2!+ · · · )
and look at the coefficient of t7, which is
x7 − 49
2x6 + 245x5 − 5145
4x4 +
11368
3− 6174x2 + 5040x− 36799
24.
This last example is different from the others in that the superscript is the same as
the subscript. Polynomials satisfying this condition are called mth order Bernoulli
polynomials of degree m. The first ten are:
B(0)0 (x) =1,
B(1)1 (x) =x− 1
2,
B(2)2 (x) =x2 − 2x +
5
6,
B(3)3 (x) =x3 − 3
4x2 + x− 3
8,
B(4)4 (x) =x4 − 8x3 + 22x2 − 24x +
251
30,
B(5)5 (x) =x5 − 5
48x4 +
35
72x3 − 25
24x2 + x− 95
288,
B(6)6 (x) =x6 − 1
40x5 +
17
96x4 − 5
8x3 +
137
120x2 − x +
19087
60480,
B(7)7 (x) =x7 − 49
2x6 + 245x5 − 5145
4x4 +
11368
3− 6174x2 + 5040x− 36799
24,
B(8)8 (x) = x8 − 1
1260x7 +
23
2160x6 − 7
90x5 +
967
2880x4 − 469
540x3 +
363
280x2 − x +
1070017
3628800,
B(9)9 (x) =x9 − 81
2x8 + 702x7 − 6804x6 +
202041
5x5 − 151389x4 + 354372x3
− 493128x2 + 362880x− 2082753
20.
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As a consequence of Definition 3, we can also formulate the equation
B(`)m (x) = (x + b1 + b2 + · · ·+ b`)
m.
The right-hand side is to be interpreted as follows: When it is expanded, each bji
occurring in the expansion is to be replaced by Bj (the jth Bernoulli number). The
above formula can be derived from Definitions 1 and 3 using the equation
∞∑m=0
B(`)m (x)
tm
m!=
(t
et − 1
)`
ext
=
(∏̀i=1
ebit
)ext
= e(x+b1+b2+···+b`)t
=∞∑
m=0
(x + b1 + b2 + · · ·+ b`)m tm
m!,
where again occurrences of bji are to be replaced by Bj.
We are interested in the irreducibility of the polynomials B(m)m (x). For this we
will use Eisenstein’s Criterion.
Definition 4. Let f(x) =∑n
i=0 aixi ∈ Z[x]. We say f(x) is Eisenstein (or
p-Eisenstein) if p - an, and p|ai for all i 6= n but p2 - a0.
Theorem 4. Eisenstein polynomials are irreducible over Q.
Proof. To prove this theorem we assume that there exist non-constant polyno-
mials g(x) and h(x) in Z[x] such that f(x) = g(x)h(x). Let g(x) =∑r
i=0 bixi and
h(x) =∑s
i=0 cixi where r + s = n. Now assume that ρ is the smallest i such that
p - bρ and that σ is the smallest i such that p - cσ. These exist because an = brcs is
not divisible by p. The coefficient of xρ+σ in f(x) = g(x)h(x) is
b0cρ+σ + b1cρ+σ−1+ · · ·+ bρ−1cσ+1 + bρcσ
+bρ+1cσ−1+ · · ·+ bρ+(σ−1)c1 + bρ+σc0 ≡ bρcσ 6≡ 0 (mod p).
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This is true because every term bicj except bρcσ is such that i < ρ j < σ. We also
know thatp|ai for all i except i = n. Hence, bρcσ = brcs = an so that |rho = r and
σ = s. But by the definition of ρ and σ we obtain p|b0 and p|c0 contradicting, that
p2 - a0. �
For our purposes, we will want the notion of an f(x) in Q[x] (not necessarily in
Z[x]) being Eisenstein (or p-Eisenstein). So suppose f(x) =∑n
i=0(ai/bi)xi where for
each i we have that ai and bi are relatively prime integers with bi > 0. Then f(x) is
Eisenstein (or p-Eisenstein) if p - bi for all i, p - an, p|ai for all i < n, and p2 - a0.
Observe that if D = lcm(b0, b1, . . . , bn) and f(x) is p-Eisenstein, then Df(x) is a p-
Eisenstein polynomial in Z[x]. Hence, f(x) in Q[x] being Eisenstein implies that it is
irreducible over the rationals.
For p a prime and m a non-negative integer, define S(m) = Sp(m) to be the sum
of the base p digits of m. Thus, if
m =n∑
j=0
ajpj
with 0 ≤ ai ≤ p− 1 for 0 ≤ i ≤ n, then
S(m) = a0 + a1 + · · ·+ an.
We also set
r :=
[m
p− 1
]and σ := m− r(p− 1).
Lemma 3. The following are equivalent:
A) S(m) < p− 1
B) S(m) = σ.
Proof. First we will show that if S(m) < p−1 then S(m) = σ. By manipulating
the definition of m we get the following:
m =n∑
j=0
ajpj
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=n∑
j=0
aj((p− 1) + 1)j
≡n∑
j=0
aj1j (mod p− 1)
≡ S(m) (mod p− 1).
On the other hand, m− (p−1) = ( mp−1
−1)(p−1) < [ mp−1
](p−1) ≤ mp−1
(p−1) = m so
that 0 ≤ σ < p−1. Also, σ ≡ m (mod p−1). Since 0 ≤ S(m) < p−1 and S(m) ≡ m
(mod p−1), we deduce S(m) = σ. Now to show that if S(m) = σ then S(m) < p−1,
we simply note that σ < p− 1 and S(m) = σ implies S(m) < p− 1. �
The following theorem from [AA] will be used but not proven in this thesis.
Theorem 5. Let ` and m be a positive integers and suppose that m < p2, that
σ > 1 and that p||`. If m is even and Bσ 6≡ 0 (mod p), then B(`)m (x) is p-Eisenstein.
We apply this theorem to the case m = ` = 2kp where k is a positive integer and
p is a prime greater than 2k + 1. First let’s see if we have the necessary conditions to
apply the theorem. Because ` = m is of the form 2kp, ` and m are positive integers so
that the first condition is satisfied. We know that 2k < p−1 so m = 2kp < (p−1)p <
p2, which implies the second condition is satisfied. Since we are considering m of the
form 2kp, with 2k < p − 1 we get S(m) = 2k, r = 2k, and σ = 2k. Thus, σ > 1.
Next, since 2k < p − 1 we know that p - 2k. Hence p||`. Note that m is obviously
even. The last condition that p - B2k is not as straight forward as the others and we
will spend the remainder of the thesis examining this property and estimating how
often it holds. To summarize, as a consequence of Theorem 5 we have shown that if
m = 2kp where k is a positive integer and p is a prime > 2k + 1 and if p - B2k, then
B(m)m (x) is p-Eisenstein. Here, p - B2k means that if B2k is expressed as a reduced
fraction N2k/D2k, both its numerator and denominator are not divisible by p.
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Since 2k < p− 1, we know (p− 1) - 2k and conclude that p - D2k because of the
von Staudt-Clausen theorem (Theorem 2). So now, with all of this in mind, we can
restate the previous theorem for our particular case as follows:
Theorem 6. If m = 2kp with k a positive integer and p a prime > 2k + 1 and if
p - N2k then B(m)m (x) is p-Eisenstein.
Now we will formulate some lemmas, in order to examine those p such that p - N2k
and try to get an estimate on how many B(m)m (x) are p-Eisenstein.
Lemma 4. The number 2(22k − 1)B2k is an integer.
Proof. By Theorem 2 and the comment following its proof, p|D2k if and only if
(p− 1)|(2k). Also, p|D2k implies p||D2k. We need only show then that if (p− 1)|2k,
then p|(2(22k − 1)
). Let p be a prime with p − 1 dividing 2k. If p = 2, then clearly
p|(2(22k − 1)
). If p 6= 2, then Fermat’s Little Theorem implies p divides 22k − 1 as
(p− 1)|(2k). Thus, p|(2(22k − 1)
). This proves the lemma. �
Lemma 5. For each positive integer k, |N2k| < (2k)2k.
Proof. First, by Theorem 1 we have
|B2k| =2(2k)!
(2π)2kζ(2k).
Since
ζ(2k) =∞∑
n=1
1
n2k
and ζ(2) = π2/6, it is easy to see that ζ(2k) ≤ ζ(2) < 10/6 < 2. We see, by Theorem
1, that
|B2k| =2(2k)!
(2π)2kζ(2k) <
4(2k)!
(2π)2k=
4(2k)!
(22k)(π2k)≤ 4(2k)!
(22k)(π2)<
4(2k)!
8(22k)<
(2k)!
2(22k).
From Lemma 4 we obtain
|N2k| ≤ 2(22k − 1)|B2k| <2(22k − 1)
2(22k)(2k)! < (2k)! < (2k)2k,
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which proves the lemma. �
We now use Theorem 5 to find a lower bound on the number of m ≤ u for which
B(m)m (x) is Eisenstein. We view u as being sufficiently large. Fix ε ∈ (0, 1/3). Set
θ = (2/3) + ε. Consider a positive integer k for which 2k < u1−θ. Let P2k denote the
set of primes dividing N2k such that
(12) uθ < p ≤ u
2k.
From Lemma 5, we obtain
(uθ)|P2k| ≤∏
p∈P2k
p ≤ |N2k| ≤ (2k)2k ≤ u(1−θ)u1−θ
.
Taking logarithms of both sides we get
(θ|P2k|) log u ≤ (1− θ)u1−θ log u =⇒ θ|P2k| ≤ (1− θ)u1−θ
=⇒ |P2k| ≤(
1− θ
θ
)u1−θ < u1−θ.
Let
P =⋃
k≤u1−θ/2
P2k.
Then P has the following properties:
(A) If p 6∈ P and (12) holds, then p - N2k.
(B) |P| ≤∑
k≤u1−θ/2
|P2k| ≤ u2−2θ.
We consider the m ≤ u of the form 2kp with p > uθ. Since θ > 2/3 and p|m,
there is exactly one such p corresponding to a given m. Also, 2kp ≤ u and p > uθ
implies p ≤ u/(2k) and 2k ≤ u1−θ < u1/3 − 1 < p − 1. In particular, (12) holds.
Also, from (A) and Theorem 6, if p 6∈ P , then B(m)m (x) is p-Eisenstein. Again noting
the uniqueness of p for a given m, we deduce that the number of m ≤ u for which
B(m)m (x) is Eisenstein is at least∑
m≤um even
∑p|m
p>uθ
p6∈P
1 =∑
uθ<p≤up6∈P
∑m≤u
(2p)|m
1.
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We give a justification for the equation above. More specifically, we show that
a pair (m, p), where m is a positive integer and p is a prime, satisfies the condition
under the double summation on the left if and only if (m, p) satisfies the conditions
under the double summation on the right. The conditions on the left we list as (i)-(v)
and on the right as (i′)-(iv′). These are as follows:
(i) m ≤ u (i′) m ≤ u
(ii) m even (ii′) uθ < p ≤ u
(iii) p|m (iii′) (2p)|m
(iv) p > uθ (iv′) p 6∈ P
(v) p 6∈ P
We begin showing that if (m, p) satisfies (i)-(v), then (m, p) satisfies (i′)-(iv′).
Obviously (i) holds if and only if (i′) holds and (v) holds if and only if (iv′) holds.
Observe that (iv) and u large imply p > 2. Hence, (ii) and (iii) imply (iii′). And since
(iii) implies p ≤ m, (i) and (iii) imply p ≤ m ≤ u. Combining this with (iv) we see
that (ii′) holds.
Now we show that if (m, p) satisfies (i′)-(iv′) then (m, p) satisfies (i)-(v). The
conditions (i) and (v) follow directly from (i′) and (iv′). Also,(ii′) implies (iv) holds,
and (iii′) implies (ii) and(iii) hold.
Define Ep by [u
2p
]=
u
2p+ Ep.
Hence, −1 < Ep ≤ 0. Making use of this notation, we obtain
∑uθ<p≤u
p6∈P
∑m≤u
(2p)|m
1 =∑
uθ<p≤up6∈P
[u
2p
]=
∑uθ<p≤u
p6∈P
(u
2p+ Ep
).
Observe that ∑uθ<p≤u
p6∈P
|Ep| ≤∑p≤u
1 � u
log u.
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Also, ∑uθ<p≤u
p∈P
u
2p≤ 1
2u1−θ|P|,
the latter being an upper bound on the number of terms times an upper bound on
the size of the terms. From (B), we deduce∑uθ<p≤u
p∈P
u
2p< u3−3θ = u1−3ε.
Using the asymptotic formula∑p≤z
1
p= log log z + C + O(1/ log z),
we obtain ∑uθ<p≤u
u
2p=
u
2
(log log u− log log uθ + O(1/ log u)
)=
u
2(log(1/θ) + O(1/ log u))
=log(1/θ)
2u + O(u/ log u).
Using that ∑uθ<p≤u
p6∈P
(u
2p+ Ep
)=
∑uθ<p≤u
u
2p−
∑uθ<p≤u
p∈P
u
2p+∑
uθ<p≤up6∈P
Ep,
we obtain that the number of m ≤ u for which B(m)m (x) is Eisenstein is at least
log(1/θ)
2u + O(u1−3ε) + O(u/ log u).
The first term dominates the above for any ε > 0. By allowing ε to approach 0 and
noting
log(3/2)
2= 0.2027...,
we deduce that, for u sufficiently large, more than one-fifth of the integers m ≤ u
are such that B(m)m (x) is Eisenstein. Hence, the Main Theorem of this chapter (and
thesis) is established.
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Page 26
Bibliography
[AA] Adelberg, Arnold, Congruence of p-adic Integer Order Bernoulli Numbers, Jour-
nal of Number Theory, 59 (1996), 374-388.
[BS] Borevich, Z.I. and Shafarevich, I.R., Number Theory, Academic Press, New York,
1966.
[HE] Edwards, Harold M., Fermat’s Last Theorem: A Genetic Introduction to Alge-
braic Number Theory, Springer-Verlag, New York, 2000.
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