Solution
= Solution
We prove the <contrapositive>. If $3\nmid a$, <Euclidean division> gives $a=3q+1$ or $a=3q+2$ for an <integer> $q$. Squaring gives respectively $a^2=3(3q^2+2q)+1$ or $a^2=3(3q^2+4q+1)+1$. Both have remainder $1$ upon division by $3$, so $3\nmid a^2$. Taking the <contrapositive> yields \b[$3\mid a^2\Rightarrow3\mid a$].