Answer
Let $n$ be some integer such that $n\ mod\ 5=3$. Then by the definition of the $mod$ function, $n=5q+3$ for some integer $q$. Squaring both sides, we get $n^{2}=(5q+3)^{2}=25q^{2}+30q+9$$=5(5q^{2}+6q)+9=$$5(5q^{2}+6q)+5+4=$$5(5q^{2}+6q+1)+4$. Since $0\leq4\lt5$, and since $q^{2}+6q+1$ is an integer, we conclude by the quotient-remainder theorem and the definition of $mod$ that $n^{2}\ mod\ 5=4$. Since $n$ was an arbitrarily chosen integer, it must be that, if $n\ mod\ 5=3$ for any integer $n$, then $n^{2}\ mod\ 5=4$.
Work Step by Step
Page 180 gives the quotient-remainder theorem, and page 181 gives the definition of the $mod$ function.