Prove that if $n\neq 2 \pmod 4$ then there exists a primitive Pythagorean triple $(x,y,z)$ in which $x$ or $y$ equals $n$

Question

If $n\neq2\mod 4$, then there are three possibilities:

$n\equiv0\pmod 4, n\equiv1\pmod 4$, and $n\equiv3\pmod 4.$

Can I have some help after this point? Thanks.

Answer 1

From $x^2+y^2=z^2$ we have $$x^2=(z-y)(z+y)$$ so your goal is to find two factors $a<b$ such that $ab=x^2$, $a=z-y$, $b=z+y$. The system $$\left\lbrace\begin{matrix}z-y=a\\z+y=b\end{matrix}\right.$$ has an integer solution if and only if $a$ and $b$ have the same parity. What happens for $x=1$?

Answer 2

since n is not of the form 4n+2, it can be written as the difference of 2 squares, say $$n = a^2 - b^2$$ we can now produce $$m = 2ab$$ so now, $$m^2 + n^2 = (a^2 + b^2)^2$$, which is a primitive pythagorean triple.