Mod 8

Claim: All perfect squares are equivalent to $0$ , $1$ , or $4$ $\text{(mod 8)}$ . Proof: If $x$ is even, then $x^2$ is divisible by $4$ , meaning it is equivalent to $0$ or $4$ $\text{(mod 8)}$ . Since $8^2 \equiv 0$ $\text{(mod 8)}$ and $2^2 \equiv 4$ $\text{(mod 8)}$ , both cases are possible.

If $x$ is odd, then $x$ can be written in the form $2y+1$ for some integer $y$ . Squaring this yields $4y^2+4y+1$ . We factor this as $4y(y+1) + 1$ . Since either $y$ or $y+1$ must be even, the first term is divisible by $8$ and the expression is equivalent to $1$ $\text{(mod 8)}$ .

Thus the possible values are ${0,1,4}$ , so there are $3$

$n:4, 5, 6,7,8,9,10,11,......$

$x:0,1,4,1,0,1,4,1,0,.......$

The remainders, $x$ , are a pattern of $0,1,4$ . So the answer is $\boxed{3}$