I have this problem that I'm a bit stuck on:
Find $m\equiv 1\pmod4$ so that $x^2\equiv -1\pmod{m}$ has no solution in $\mathbb{Z}$.
So far, I know that $m$ can't be prime because $(\frac{-1}{p})=1$, $p$ prime, whenever $p\equiv 1\pmod4$, where $(\frac{}{})$ is the legendre symbol.
Also, I've considered the following: $m=4k+1$, so
$(\frac{-1}{m})=(\frac{4k}{m})=(\frac{4}{m})(\frac{k}{m})=(\frac{k}{m})$, but again I get stuck here since $m$ is composite.
I think that there is no such $m$, but I'm unsure how to prove it.
My Proof So Far
Note that if $m=p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$, $p_i$ prime, $k_i\geq 1$ then by the Chinese Remainder Theorem, $x^2\equiv -1\pmod{m}$ has a solution $\iff x^2\equiv -1\pmod{p_i^{k_i}}$ for all $1\leq i\leq r$.
So, let us take $m=p^k$, $p$ prime, $k\geq 1$.
Note that we must have $k>1$ since $k=1\implies m=p\implies (\frac{-1}{m})=1$ since $m\equiv1\pmod4$.
...
Now, I know that $m=3^2=9$ works, but I'm not sure how to prove that $({-1\over9})=-1$ since $9$ is composite. I'm also not allowed to use the fact that $x^2\equiv a\pmod{p^n}$ has a solution $\iff ({a\over p})=1$ since we haven't covered this theorem in class.