Suppose $a = 2k$, for some integer k, and $n = a^2=4k^2$ which is zero mod 4. Otherwise, a is odd, and $a = 2k+1$. Then $n = a^2=(2k+1)^2=4k^2+4k+1= 4(k^2+k)+1$. This is 1 mod 4. This is desired result.
In general, the contrapositive of a conditional:
'If P then Q'
is the statement:
'If not Q then not P'
Applied to your statement, we would thus get:
'If $n$ is a perfect square, then $n$ is not a positive integer such that $n \equiv 2 \pmod{4}$ or $n \equiv 3 \pmod{4}$'
... But somehow I doubt that's what they meant. In fact, the original statement was probably meant as:
'For any positive integer $n$, it holds that if $n \equiv 2 \pmod{4}$ or $n \equiv 3 \pmod{4}$, then $n$ is not a perfect square'
So then by taking the contrapositive of the contrapositive of the conditional that is part of that general statement about positive integers, we get:
'For any positive integer $n$, it holds that if $n$ is a perfect square, then it is not the case that $n \equiv 2 \pmod{4}$ or $n \equiv 3 \pmod{4}$'
...which makes a lot more sense.
Indeed, to prove this statement:
Take $n$ to be a positive integer and assume it is a perfect square. So, $n=k^2$ with $k$ an integer. $k$ is either even or odd. If $k$ is even, then $k=2m$ for some integer $m$, and so $n=(2m)^2=4m^2$. Hence, $n \equiv 0 \pmod{4}$. If $k$ is odd, then $k=2k+1$ for some integer $m$' and so $n=(2m+1)^2=4m^2+4m+1$, and hence $n \equiv 1 \pmod{4}$. So, it is not the case that $n \equiv 2 \pmod{4}$ or that $n \equiv 3 \pmod{4}$.
Best Answer
$8\mod 3=2$ but $8$ is not a...