hikorski triples by jonny griffiths uea, may 2010
TRANSCRIPT
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Hikorski Triples
By Jonny GriffithsUEA, May 2010
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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
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Mathematics
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What does
mean to you?
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GCSE Resit Worksheet, 2002
How many different equations can you make by putting the numbers into the circles?
Solve them!
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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.
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This maximum is possible: for example, if 7, -2, 3 and 4 are in the bag,
then the solutions are:
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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)
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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
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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.
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Are (7, 5, 3) linked in any way?
Will this always work?
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a, b, c, d in the bag gives the same as
b, c, d, a in the bag, gives the same as …
Permutation Law
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a, b, c, d in the bag gives the same as
a + k, b + k, c + k, d + kin the bag.
Translation Law
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a, b, c, d in the bag gives the same as
ka, kb, kc, kdin the bag.
Dilation Law
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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.
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a, b, c, d in the bag gives the same as
-a, -b, -c, -din the bag.
Reflection Law
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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
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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.
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Given a bag of numbers in Standard Form,
where might the whole numbers for our HT come from?
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The only possible whole numbers here are:
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(b-1)/a must be the smallest here.
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Pythagorean Triples(√(x2+y2), x, y)
Hikorski Triples(p, q, (pq+1)/(p+q))
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How many HTs are there?
Plenty...
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Is abc unique?
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Twelve solutions to bag problem are:
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What do
mean to you?
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Adding Speeds Relativistically
Suppose we say the speed of light is 1.
How do we add two speeds?
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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...
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But try the recurrence relation:x, y, -(xy+1)/(x+y)…
Periodic, period 3
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Now try the recurrence relation:x, y, -(x+y)/(xy+1)…
Also periodic, period 3
Are both periodic, period 6
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Parametrisation for Pythagorean Triples:
(r(p2+q2), 2rpq, r(p2-q2))
For Hikorski Triples?
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The Cross-ratio
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If a, b, c and d are complex,
when is the cross-ratio real?
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Takes six values as A, B and C permute:
Form a group isomorphic to S3 under composition
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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.
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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
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So we have six isomorphic elliptic curves.
The j-invariant for each will be the same.
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is an elliptic curve
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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)
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x y
Find a in terms of x and e in terms of y and then substitute...
Cross-ratio-type functions and Lyness Cycles
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What if we try the same trick here?
)(
)(
dc
ba
)(
)(
ed
cb
)(
)(
ae
dc
)(
)(
ba
ed
x
y
z
?
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And here?
)(
)(
bc
ba
)(
)(
cd
cb
)(
)(
da
dc
x
y
?
So this works with the other cross-ratio type functions too...
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Why the name?
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www.jonny-griffiths.net