Only $p=3$, which clearly works. Note that if the sets are equal modulo $p$, then $\sum_{a\in A}a^2\equiv \sum_{b\in B}b^2\pmod{p}$. We next compute each of these sums. On the one hand \[ \sum_{1\le k\le \frac{p-1}{2}} \left(k^2+1\right)^2 = \frac{p-1}{2} + \sum_{1\le k\le \frac{p-1}{2}} k^4 + \sum_{1\le k\le \frac{p-1}{2}}k^2\equiv \frac{p-1}{2}\pmod{p}, \]unless $p\in\{3,5\}$ (and $p=5$ does not work), using the identities \[ \sum_{1\le k\le n}k^2 = \frac{n(n+1)(2n+1)}{6}\quad\text{and}\quad \sum_{1\le k\le n}k^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}. \]Notice, on the other hand \[ \sum_{1\le m \frac{p-1}{2}}g^m = g^2+g^4+\cdots+g^{p-1} = g^2 \cdot \frac{g^{p-1}-1}{g^2-1} \equiv 0\pmod{p}, \]using $p\mid g^{p-1}-1$ and the fact $g^2\not\equiv 1\pmod{p}$. This completes the proof.