Representations of an odd integer correspond to representations of twice that integer

If $x^2+y^2=m$, oddness of $m$ makes $x,y$ have opposite parity, so $x+y$ and $x-y$ are odd; moreover $(x+y{)}^2+(x-y{)}^2=2m$, so $\mathrm{\Phi }$ sends representations of $m$ to representations of $2m$.

If $u^2+v^2=2m$, reduction modulo $4$ shows that $u$ and $v$ are both odd: their squares cannot both be even because $2m\equiv 2(\mathrm{m}\mathrm{o}\mathrm{d}4)$, and they cannot have opposite parity because their square sum would be odd. Thus $\mathrm{\Psi }(u,v)$ has integer coordinates, and their squared sum is $(u^2+v^2)\mathrm{/}2=m$.

Direct substitution gives $\mathrm{\Psi }(\mathrm{\Phi }(x,y))=(x,y)$ and $\mathrm{\Phi }(\mathrm{\Psi }(u,v))=(u,v)$.

Suppose $(x,y)$ is primitive and a common divisor greater than $1$ divides both coordinates of $\mathrm{\Phi }(x,y)$. By [L1] it has a prime divisor $\mathrm{\ell }$. Since $x+y$ and $x-y$ are odd, $\mathrm{\ell }$ is odd; because $\mathrm{\ell }$ divides $2x$ and $2y$, [L2] gives $\mathrm{\ell }\mid x$ and $\mathrm{\ell }\mid y$, contradicting primitivity. Hence $\mathrm{\Phi }(x,y)$ is primitive.

Suppose $(u,v)$ is primitive and a common divisor greater than $1$ divides both coordinates of $\mathrm{\Psi }(u,v)$. A prime divisor $\mathrm{\ell }$ supplied by [L1] then divides their sum $u$ and difference $v$, contradicting primitivity. Hence $\mathrm{\Psi }(u,v)$ is primitive.

The mutually inverse maps of step 2.1 give the first bijection, and steps 2.2 and 2.3 show that they restrict to mutually inverse maps on primitive representations.