Finding the number of n, the creator of the perfect square

Plug in 1 to see that 1 does work.

Now let's assume that $n > 1$ .

Taking the expression mod 3, we have $(-1)^n + 1 \equiv p^2$

This implies that $n$ must be odd since $p^2$ in mod 3 can only be 0 or 1.

Taking the expression mod 4, we have $2^n + (-1)^n \equiv (-1)^n \equiv p^2$

This implies that $n$ must be even since $p^2$ in mod 4 can only be 0 or 1.

However this is a contradiction which implies that if $n > 1$ , there are no values.

Therefore the only value is 1, implying that there is only $\boxed{1}$ value.

2011 USAJMO #1