Prove that if x² + y² = z² with x,y,z integer that a p and q integer can be found such that
x=2pq , z = p² + q² and y = p² - q²

We can assume that (x,y,z) = 1 because if not the common factors can be divided out.

So (x,y)=1 , (x,z)=1 and (y,z)=1.

Now because x and y has no common factors , both cannot be even.
So either both be odd or one even and the other odd.

Let us now assume that both x and y is odd. If so then z is even. So z = 2n , z² = 4n²

Therefore 4 divides z². If this is so then 4 must also divide x² + y²

Because x odd we have x == (1 or 3)mod4 , so x² ==(1 or 9)mod4 , so x² ==1mod4

Because y odd we also have y²==1mod4

Therefore x² + y² == 2 mod 4 , so 4 does not divide x² + y² and we have
a contradiction and our assumption that both x and y is odd must be wrong.

So x is even and y is odd.

We can therefore write x² = (z-y)(z+y)

We see that x is even. Now if z-y and z+y has common factors then
so must their sums and differences also have the same common factors.

The sum is 2z and the difference is 2y. But (z,y)=1 so the only common factor is 2.

We can therefore write z+y = 2r and z-y=2s where (r,s)=1

So x² = 4rs , but x² is square and the product of r and s must therefore also be square.
However because r and s have no common factors they themselves must be squares.

So we have r = p² and s = q²

Therefore x² = 4 p²q² , so x = 2pq

We also have that 2z = 2r + 2s , so z = r + s = p² + q²

In the same way we get 2y = 2r - 2s , so y = r -s = p² - q²

Done.

Example : 4² + 3² = 5²

So 2r = 8 and 2s = 2

So r = 4 and p=2

So s=1 and q=1

This gives x = 4 = 2pq = 2x2x1

Also z = p² + q² = 2² + 1² = 4+1 = 5

Also y = p² - q² = 2² - 1² = 4-1 = 3

End