Originally Posted by

**kingwinner** __Theorem: __The positive primitive solutions of $\displaystyle x^2 + y^2 = z^2$ with y even are $\displaystyle x = r^2 - s^2, y = 2rs, z = r^2 + s^2$, where r and s are arbitrary integers of opposite parity with r>s>0 and (r,s)=1.

**Using this theorem, find all solutions of the equation $\displaystyle x^2 + y^2 = 2z^2$**

**(hint: write the equation in the form $\displaystyle (x+y)^2 + (x-y)^2 = (2z)^2$)**

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I am not sure how to do this problem. I don't even understand how to get the equation in the hint.

I hope someone can help me out. Thank you!

[also under discussion in math links forum]