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        riddles >> easy >> Triple Pythagoras
        
(Message started by: cool_joh on Dec 27^th, 2007, 5:51am)

Title: Triple Pythagoras
Post by cool_joh on Dec 27^th, 2007, 5:51am

How do you prove, or disprove, that the only solution for x²+y²=z² is x=2ab, y=a²-b² and z=a²+b² (x and y may be reversed)?

Title: Re: Triple Pythagoras
Post by JocK on Dec 28^th, 2007, 8:31am

on 12/27/07 at 05:51:41, cool_joh wrote:

How do you prove, or disprove, that the only solution for x²+y²=z² is x=2ab, y=a²-b² and z=a²+b² (x and y may be reversed)?

This can be proved for integer a, b, x, y and z by considering the fact that the (a, b)'s form the spinor space of the (x, y, z)'s with appropriate Clifford algebra.

Title: Re: Triple Pythagoras
Post by cool_joh on Dec 28^th, 2007, 6:29pm

Sorry, I was wrong. That's only the primitive solutions.

x=2abc, y=c(a^2-b^2) and z=c(a^2+b^2)

Title: Re: Triple Pythagoras
Post by JocK on Dec 29^th, 2007, 11:54am

It is better to consider the primitive triplets (adding an integer scaling factor - as you did - will then take care of the non-primitive triplets).

I was myself not very precise either. Let me restate:

There is a 1:1 mapping between the space of the primitive triplets (x, y, z) and its Cl(2,1) spinor space (a, b), provided GCD(a, b) = 1, a+b=odd and a, b =/= 0. (Note that a, b, x and y are not restricted to the natural numbers but run through all nonzero integers. )

A reference: http://www.math.siu.edu/kocik/pracki/44Cliffpdf.pdf

Title: Re: Triple Pythagoras
Post by Icarus on Jan 4^th, 2008, 7:00pm

Something hopefully a little more accessible:

If x, y, z are a primitive solution of x² + y² = z², then they are pairwise relatively prime, since any common factor of two would also have to be a factor of the third. Therefore only one of the three can be even. If any two are odd, then the third must be even since the sum or difference of odd numbers is even.

If z were even, then both x and y are odd, and x² = 4j+1 and y² = 4k+1 for some j and k. But then z² = 4(j+k) + 2 and so would not be divisible by 4, which cannot be.

Hence one of x or y is even. Since x and y are interchangeable in their roles, we can assume x is even, and y is odd.

Rewrite the equation: x² = z²-y² = (z+y)(z-y). z+y and z-y are obviously both even (since y and z are both odd). At most, only one of z+y and z-y is divisible by 4. For if 4 divided both, then 4 would also divide their sum 2z, and 2 would divide z, which we know is odd. If p is a common odd divisor of both z+y and z-y, then p divides x and p divides (z+y)+(z-y) = 2z. Since p is odd, it must be that p divides z. Since x and z are relatively prime, p can only be 1.

Let m = (z+y)/2, n=(z-y)/2. Then m and n are relatively prime, and mn = (x/2)². This requires that m and n themselves are perfect squares. Let a = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifm, b = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gifn. Then we see that x = 2ab, y = a² - b² and z = a² + b².

QED.