Hikorski Triples

By Jonny GriffithsUEA, May 2010

The mathematician's patterns, like the painter's or the poet's

must be beautiful; the ideas, like the colors or the words

must fit together in a harmonious way. Beauty is the first test:

there is no permanent place in this world for ugly mathematics.

G. H. Hardy (1877 - 1947), A Mathematician's Apology

Mathematics

What does

mean to you?

GCSE Resit Worksheet, 2002

How many different equations can you make by putting the numbers into the circles?

Solve them!

Suppose a, b, c, and d are in the bag.

If ax + b = cx + d, then the solution to this equation is x =  

There are 24 possible equations, but they occur in pairs, for example:

 ax + b = cx + d and cx + d = ax + b

will have the same solution.

 So there are a maximum of twelve distinct solutions.

This maximum is possible: for example, if 7, -2, 3 and 4 are in the bag,

then the solutions are:

If x is a solution, then –x, 1/x and -1/x will also be solutions.

ax + b = cx + d

a + b(1/x) = c + d(1/x)

c(-x) + b = a(-x) + d

a + d(-1/x) = c + b(-1/x)

The solutions in general will be:

{p, -p, 1/p, -1/p}{q, -q, 1/q, -1/q}

and {r, -r, 1/r, -1/r}

where p, q and r are all ≥ 1

It is possible for p, q and r to be positive integers.

For example, 1, 2, 3 and 8 in the bag give (p, q, r) = (7, 5, 3).

In this case, they form a Hikorski Triple.

Are (7, 5, 3) linked in any way?

Will this always work?

a, b, c, d in the bag gives the same as

b, c, d, a in the bag, gives the same as …

Permutation Law

a, b, c, d in the bag gives the same as

a + k, b + k, c + k, d + kin the bag.

Translation Law

a, b, c, d in the bag gives the same as

ka, kb, kc, kdin the bag.

Dilation Law

So we can start with 0, 1, a and b (a, b rational numbers

with 0 < 1 < a < b)in the bag without loss of

generality.

a, b, c, d in the bag gives the same as

-a, -b, -c, -din the bag.

Reflection Law

Suppose we have 0, 1, a, bin the bag, with 0 < 1 < a < b

and with b – a < 1

then this gives the same as –b, -a, -1, 0

which gives the same as 0, b - a, b - 1, b

which gives the same as 0, 1, (b -1)/(b - a), b/(b - a)

Now b/(b - a) - (b -1)/(b - a) = 1/(b - a) > 1

If the four numbers in the bag are given as {0, 1, a, b}

with 1< a < b and b – a > 1, then we can say the bag is in Standard Form.

 So our four-numbers-in-a-bag situation

obeys four laws:

the Permutation Law, the Translation Law, the Reflection Law and the Dilation Law.

Given a bag of numbers in Standard Form,

where might the whole numbers for our HT come from?

The only possible whole numbers here are:

(b-1)/a must be the smallest here.

Pythagorean Triples(√(x2+y2), x, y)

Hikorski Triples(p, q, (pq+1)/(p+q))

How many HTs are there?

Plenty...

Is abc unique?

Twelve solutions to bag problem are:

What do

mean to you?

Adding Speeds Relativistically

Suppose we say the speed of light is 1.

How do we add two speeds?

Try the recurrence relation:x, y, (xy+1)/(x+y)…

Not much to report...

Try the recurrence relation:x, y, (x+y)/(xy+1)…

still nothing to report...

But try the recurrence relation:x, y, -(xy+1)/(x+y)…

Periodic, period 3

Now try the recurrence relation:x, y, -(x+y)/(xy+1)…

Also periodic, period 3

Are both periodic, period 6

Parametrisation for Pythagorean Triples:

(r(p2+q2), 2rpq, r(p2-q2))

For Hikorski Triples?

The Cross-ratio

If a, b, c and d are complex,

when is the cross-ratio real?

Takes six values as A, B and C permute:

Form a group isomorphic to S3 under composition

 So the cross-ratio

and these cross-ratio-type functionsall obey the four laws:

the Permutation Law, the Translation Law, the Reflection Law and the Dilation Law.

Elliptic Curve Connection

Rewrite this as Y2 = X(X -1)(X - D)

Transformation to be used is:

Y = ky, X = (x-a)/(b-a), or...

D runs through the values

So we have six isomorphic elliptic curves.

The j-invariant for each will be the same.

is an elliptic curve

has integral points (5,3), (3,-2), (-2,5)

If the uniqueness conjecture is true...

and (30,1), (1,-1), (-1,30)

and (1,30), (30,-1), (-1,1)

x y

Find a in terms of x and e in terms of y and then substitute...

Cross-ratio-type functions and Lyness Cycles

What if we try the same trick here?

)(

)(

dc

ba

)(

)(

ed

cb

)(

)(

ae

dc

)(

)(

ba

ed

x

y

z

?

And here?

)(

)(

bc

ba

)(

)(

cd

cb

)(

)(

da

dc

x

y

?

So this works with the other cross-ratio type functions too...

Why the name?

www.jonny-griffiths.net