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 $\mathbb{Z}_p$?
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1$\begingroup$ Why is this equivalent? $\endgroup$– Will SawinCommented Jun 1, 2023 at 13:26
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$\begingroup$ Sorry. For the equivalent problem, I am assuming -1 is not a square in $\mathbb{Z}_p$, i.e., $p \equiv 3 \pmod 4$. That's the case I am the most interested in, but I believe the first statement holds for all $p > 3$. $\endgroup$– DomCommented Jun 1, 2023 at 13:31
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$\begingroup$ I edited the question to clarify. $\endgroup$– DomCommented Jun 1, 2023 at 13:37
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1$\begingroup$ I suppose "${\mathbb Z}_p$" is being used here for the finite field of $p$ elements, not the $p$-adic numbers. $\endgroup$– Noam D. ElkiesCommented Jun 1, 2023 at 14:15
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$\begingroup$ @NoamD.Elkies Yes, the finite field of $p$ elements. $\endgroup$– DomCommented Jun 1, 2023 at 14:32
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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
answered Jun 1, 2023 at 14:27
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$\begingroup$ Thank you! More generally, can we use a similar argument if we replace the cubic power by a larger power, say (1+x^2)^t-1? Would we have a growing number of cases to verify manually? The cases t = (p-1)/2 and t = p-1 are special and can be resolved directly. When $t \ne (p-1)/2, p-1$, I always find a lot of solutions for any p. $\endgroup$– DomCommented Jun 1, 2023 at 15:02
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2$\begingroup$ Yes, if $d$ is the degree of an absolutely irreducible curve with $N$ points in $\mathbb F_p^2$, then $\lvert N-p-1\rvert\le (d-1)(d-2)\sqrt{p}+d$, see e.g. Theorem 5.4.1 in the book "Field Arithmetic" by Fried and Jarden. $\endgroup$ Commented Jun 1, 2023 at 15:27