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Topic: Triple Pythagoras (Read 1200 times) |
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cool_joh
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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)?
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JocK
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Re: Triple Pythagoras
« Reply #1 on: Dec 28th, 2007, 8:31am » |
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on Dec 27th, 2007, 5:51am, 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)? |
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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.
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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cool_joh
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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)
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JocK
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Re: Triple Pythagoras
« Reply #3 on: Dec 29th, 2007, 11:54am » |
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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
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Icarus
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Re: Triple Pythagoras
« Reply #4 on: Jan 4th, 2008, 7:00pm » |
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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 =  m, b = n. Then we see that x = 2ab, y = a2 - b2 and z = a2 + b2.
QED.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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