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Currently learning about primitive Pythagorean triples and I'm having trouble approaching the following proof.

Given that $x, y, z$, is a primitive Pythagorean triple with $y$ even, I need to show:

$$x + y \equiv x - y \equiv 1,7 \pmod 8$$

So far all I've noticed is that $x+y$ and $x-y$ must be odd but I'm not sure how to determine the equivalence to each other $\bmod 8$. Could anyone give me a small push in the right direction?

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  • $\begingroup$ $x^2 - y^2 \equiv x^2 + y^2 \pmod 8$ since $2y^2 \equiv 0 \pmod 8 $. Since $x^2 + y^2$ is an odd square, it has to be $1 \pmod 8$. $\endgroup$
    – Callus - Reinstate Monica
    Apr 23, 2015 at 2:19

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A start: Note that $x$ and $z$ are odd, so $x^2\equiv 1\pmod{8}$ and $z^2\equiv 1\pmod 8$. It follows that $z^2-x^2\equiv 0\pmod{8}$. Thus $y^2$ is divisible by $8$, and therefore $y$ is divisible by $4$.

Now we can show that $x+y\equiv x-y\pmod{8}$. For $(x+y)-(x-y)=2y$, and since $y$ is divisible by $4$, it follows that $2y$ is divisible by $8$.

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