# Parameterization of a pythagorean triple Watch

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How can I show that the parametrization of x^2 + y^2 = z^2 gives x=2mn, y=m^2 - n^2 and z=m^2 + n^2 ? (Note: this is not equivalent to showing that one of x and y must be odd and the other even.. that can be done easily using modulo arguments. I think this is relatively harder.)

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#2

wha huh, isn't it just a case of substituting in x/y/z and showing it to be consistent???? => x^2 + y^2 = z^2 can be written as the above parameters.

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Say x^2 + y^2 = z^2 and that x,y,z have no common factor between the three of them.

Then, as you've said, one of x and y, needs to be even (choose x) and lets take the other two to be odd.

x^2 = z^2 - y^2 = (z-y)(z+y).

Let's take a prime p>2 which is a factor of z-y but which isn't a repeated factor. Then p divides x^2 and so divides x. So it's repeated prime factor of x^2. Then, as a repeated prime factor of (z-y)(z+y), it must also divide z+y.

Now we're in the case that p divides each of x, z-y,z+y and hence also x,2y,2z. As p>2 then p divides each of x,y,z - a contradiction.

So every factor of z-y and z+y other than 2 must be repeated.

We know each of x,z-y,z+y is even. Note that x^2 has an even number of factors of 2. If z-y and z+y both had even numbers of 2-factors then we'd be back to x,y,z having a common factor.

Thus we may write z-y= 2n^2 for some m and z+y = 2m^2 for some n.

Substituting in for y and z gives x = 2mn.

Then, as you've said, one of x and y, needs to be even (choose x) and lets take the other two to be odd.

x^2 = z^2 - y^2 = (z-y)(z+y).

Let's take a prime p>2 which is a factor of z-y but which isn't a repeated factor. Then p divides x^2 and so divides x. So it's repeated prime factor of x^2. Then, as a repeated prime factor of (z-y)(z+y), it must also divide z+y.

Now we're in the case that p divides each of x, z-y,z+y and hence also x,2y,2z. As p>2 then p divides each of x,y,z - a contradiction.

So every factor of z-y and z+y other than 2 must be repeated.

We know each of x,z-y,z+y is even. Note that x^2 has an even number of factors of 2. If z-y and z+y both had even numbers of 2-factors then we'd be back to x,y,z having a common factor.

Thus we may write z-y= 2n^2 for some m and z+y = 2m^2 for some n.

Substituting in for y and z gives x = 2mn.

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Thanks for the reply, Riche. I might be missing something, though. Isn't y supposed to be m^2 - n^2 and z=m^2 + n^2? Thanks again.

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#5

(Original post by

Thanks for the reply, Riche. I might be missing something, though. Isn't y supposed to be m^2 - n^2 and z=m^2 + n^2? Thanks again.

**J.F.N**)Thanks for the reply, Riche. I might be missing something, though. Isn't y supposed to be m^2 - n^2 and z=m^2 + n^2? Thanks again.

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