Solution
= Solution
The <contrapositive> is that an <integer> <perfect square> has residue $0$ or $1$ modulo $4$. Let $n=t^2$ for an <integer> $t$. If $t=2u$ is an <even number>, then $t^2=4u^2\equiv0\pmod4$. If $t=2u+1$ is an <odd number>, then $t^2=4u(u+1)+1\equiv1\pmod4$. These exhaust the possible parities of an <integer>. Thus a residue of $2$ or $3$ excludes being a <perfect square>, as required by the <contrapositive>.