Past exam of the year 1 of the computer science course of the University of Oxford / 2021 / Discrete mathematics, continuous mathematics, and probability / 2 / b / ii / 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 $t^2=4u^2\equiv 0(\mathrm{mod}4)$. If $t=2u+1$ is an odd number, then $t^2=4u(u+1)+1\equiv 1(\mathrm{mod}4)$. 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.