Prove that $x + y \equiv x − y \equiv 1 \text{ or } 7 \mod 8$ where $x, y, z$ is primitive Pythagorean Triple. 
Let $x, y, z$ be a primitive Pythagorean triple where $y$ is even.
I'm trying to show that $x + y  \equiv x − y \equiv 1 \text{ or } 7 \mod 8$.

What I have so far is the following:
$x \equiv 1 \mod 2$ and $y \equiv 0 \mod 2$
So, $$x+y\equiv 1\mod 2 \\x-y\equiv 1 \mod 2$$
Since $x+y$ and $x-y$ are odd, they can be equivalent to $1, 3, 5, 7 \mod 8$.
I have no idea where to go from here, or even if the last step is even correct/moving me in the right direction.
 A: There are integers $u>v>0$ such that $x=u^2-v^2$ and $y=2uv$. Furthermore, $u$ and $v$ are coprime and have different parity. This shows that $y$ is a multiple of $4$, and therefore, $x-y\equiv x+y\pmod 8$.
On the other hand, since $z$ is odd and $y$ is a multiple of $4$,
$$(x+y)^2=z^2+2xy\equiv 1\pmod 8$$
so $x+y\equiv\pm1\pmod 8$.
A: OK, so the first thing to see is what a primitive Pythagorean triple is.
That's where $x, y, z$ are coprime. 
So, given that, both $x$ and $z$ cannot be even, but one or the other might work.
To look at that, we see that $y^2$ must be even, and $z^2$ being even would force $x^2$ to be even. Therefore, we can see that neither $x$ nor $z$ can be even.
To find a pythagorean triple given that the first term is odd and the next term is even.
I am pretty sure that for all odd numbers, there is only one given primitive pythagorean triple, which is given by the form $x, floor(x^2/2), ceiling (x^2/2)$
All odd numbers are either 1, 3, 5, or 7 mod 8. Therefore, using the formula, we would get that the respective pythagorean triples would be 0, 4, 4, or 0, and their sums would be therefore, the sums would be 1, 7, 1, and 7; the differences would be 1, -1 (or 7), 1, and 7
