jmc

The prime factorization of $120$ is $120=2^3\cdot 3\cdot 5$. By the Chinese Remainder Theorem, it suffices to evaluate all possible remainders of $p^2$ upon division by each of $2^3$, $3$, and $5$. Since $p$ must be odd, it follows that $p=2k+1$ for some integer $k$. Thus, $(2k+1{)}^2=4k^2+4k+1=4(k)(k+1)+1$, and since at least one of $k$ and $k+1$ is even, then $$p^2\equiv 8\cdot \frac{k(k+1)}{2}+1\equiv 1(\mathrm{mod}8).$$Since $p$ is not divisible by $3$, then $p=3l\pm 1$ for some integer $l$, and it follows that $$p^2\equiv (3k\pm 1{)}^2\equiv (\pm 1{)}^2\equiv 1(\mathrm{mod}3).$$Finally, since $p$ is not divisible by $5$, then $p=5m\pm 1$ or $p=5m\pm 2$ for some integer $m$. Thus, $$p^2\equiv (5k\pm 1{)}^2\equiv 1(\mathrm{mod}5)\text{ or }{p}^2\equiv (5k\pm 2{)}^2\equiv 4(\mathrm{mod}5).$$We now have two systems of three linear congruences; by the Chinese Remainder Theorem, there are exactly $\boxed{2}$ remainders that $p^2$ can leave upon division by $120$. We can actually solve the congruences to find that $p^2\equiv 1,49(\mathrm{mod}120)$: for $p=7$, we have $p^2=49$, and for $p=11$, we have $p^2=121\equiv 1(\mathrm{mod}120)$.