Show that if $m^2 + n^2 $ is divisible by $4$, then $mn$ is also divisible by $4$. Show that if $m$ and $n$ are integers such that $m^2 + n^2 $ is divisible by $4$, then $mn$ is also divisible by $4$.
I am not sure where to begin. 
 A: Note that $n^2$ is even if and only if $n$ is even.
Since $m^2+n^2$ is even, you see that either both $m,n$ are even, or both are odd.
If both $m$, $n$ are even, then $mn$ is divisible by $4$.
If both $m$, $n$ are odd, you have $m=2k+1$, $n=2l+1$ so $m^2+n^2=4k^2+4k+4l^2+4l+2$ is not divisible by $4$. This gives the result.
A: $(m+n)^2=m^2+n^2+2mn$,  let $m^2+n^2=4k$ then $(m+n)^2=4k+2mn$ , now as RHS is divisible by $2$ LHS will also be divisible by $2$ but LHS is square of some quantity thus if $2$ divides it $4$ will also divide now consider $(m+n)^2-4k=2mn$. Now LHS is divisible by $4$ so $2mn$ is divisible by $4$ thus $mn$ is divisible by $2\implies$ atleast one of $m$ or $n$ is even but as $m^2+n^2$ is even , both of them must be even and therefore $mn$ is divisible by $4$. 
A: The worst case is that they are both $2\pmod 4$, hence their product is $0\pmod 4$. For if they are both $0\pmod 4$ then their product is readily $0\pmod 4$.
A: A square is congruent to $0$ or $1$ modulo $4$. Hence if $\,\,m^2+n^2\equiv 0 \mod 4$, $\,\,m^2\equiv 0\equiv n^2\mod 4$, so $\,\,m,n\equiv 0\mod 2 $. We conclude that $\,\,m\cdot n \equiv 0\mod 4$.
