Let $p > 3$ be prime. Is is true that there exists $x \in \mathbb{Z}\_p$ such that
$$
(1+x^2)^3-1
$$
is not a square in $\mathbb{Z}\_p$? In particular, when $-1$ is not a square in $\mathbb{Z}\_p$, can we show that the equation
$$
-y^2 = (1+x^2)^3-1
$$
always has non-trivial solutions $(x,y) \ne (0,0)$ in $\mathbb{Z}\_p$?



Polynomial that is not always a square over $\mathbb{Z}_p$
Yes, this is true.


We have $(1+x^2)^3-1 = x^2 (x^4 + 3x^2 + 3)$, so we're asking for
nonzero x such that $x^4 + 3x^2 + 3$ is not a square. If none exist then
the genus-1 curve $y^2 = x^4 + 3x^2 + 3$ has at least $2p-4$ rational points
over the $p$-element field (including the two points at infinity, and
subtracting at most 4 for the zeros of $x^4 + 3x^2 + 3 \bmod p$).
We can now use the Hasse bound, which says that the number of points is
at most $p + 2\sqrt{p} + 1$, to give an upper bound on $p$; explicitly,
if $2p-4 \leq p + 2\sqrt{p} + 1$ then $p < 7 + \sqrt{24} < 12$.
So we need only only exhibit solutions for $p=5,7,11$,
and it turns out that $x = \pm 4$ works for each of them. **QED**


