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The question is,

Show that $$\sum_{x \in \mathbb{F}_p} \left(\frac{x^2-1}{p}\right) = -1$$ where the operation $\left(\frac{x}{p}\right) = \pm 1$ if $x$ is a quadratic residue/non-residue and $0$ otherwise.

I was able to follow the argument given here Prove that $\sum_{X=0}^{p-1} \left(\frac{X^{2}+A}{p}\right)=-1$ for the $\left(\frac{-1}{p}\right) = -1$ case, but I'm not sure how to make progress on the other case (i.e $p\equiv 1$ mod $4$).

One thought I had that I couldn't quite make work was to try and use the formula $\left(\frac{x}{p}\right) \equiv x^r$ mod $p$, where $r = (p-1)/2$ and then expanding using the binomial theorem.

Any thoughts or hints would be greatly appreciated!

user413766
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  • Hint: It's enough to know how many $x \in \mathbb{F}_p$ exist for which $x^2 - 1$ is a square in $\mathbb{F}_p$. How can you rewrite the equation $x^2 - 1 = y^2$ in a nicer form? – darij grinberg Oct 02 '18 at 21:01
  • I'm not quite sure what you're getting at, but I know we can write $1 = x^2-y^2 = (x+y)(x-y)$ giving us that $(x+y)$ and $(x-y)$ are inverses mod $p$. I'm not sure what else that would give us though. – user413766 Oct 02 '18 at 22:39
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    Correct. Try solving that equation in $x+y$ and $x-y$ instead of $x$ and $y$. – darij grinberg Oct 02 '18 at 22:39
  • Which equation? Do you mean re-writing the summation in terms of $x+y$ and $x-y$ somehow? I mean, either they're both residues or both non-residues, but does that help us? – user413766 Oct 02 '18 at 23:06
  • Do you see that it suffices to count the pairs $\left(x,y\right)$ satisfying $1 = \left(x+y\right)\left(x-y\right)$ (rather than merely the $x$'s involved in these pairs)? Then, you should see why this boils down to counting the pairs $\left(a,b\right)$ satisfying $1=ab$. – darij grinberg Oct 03 '18 at 02:42
  • I see that there should be exactly $p-1$ pairs $(a,b) \in \mathbb{F}_p^2$ that have the property $1 \equiv ab$, but I'm missing the connection to how this applies to our sum. – user413766 Oct 03 '18 at 20:30
  • Let us continue this discussion in chat. – user413766 Oct 03 '18 at 20:38
  • I would love any thoughts you might have about the solution I posted to this problem – user413766 Oct 04 '18 at 03:25

1 Answers1

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This is my attempt at what I believe is a full answer following the hints given in the comments.

We first show that there are exactly $p-1$ solutions to the equation $$y^2 = x^2 - 1$$ for pairs $(x,y)\in \mathbb{F}_p^2$. If $y^2 = x^2 -1$ then $(x-y)(x+y) = 1$. Letting $$u = x-y$$ $$v = x+y$$ we see that for any choice of $(u,v) \in \mathbb{F}_p^2$ we can uniquely solve the system of equations for $x$ and $y$. Thus the number of solutions to $1 = (x-y)(x+y)$ is just the number of solutions to $1 = uv$. There are $p-1$ choices for $u$, and once $u$ is chosen, there is only one $v$ for which $1 = uv$. We conclude that there are exactly $p-1$ solutions to $y^2 = x^2 -1$

We will use this information to count the number of distinct $x$ values for which $(\frac{x^2-1}{p}) = 1$. We have the two solutions $(\pm1,0)$, and then for each other value of $x$ for which $y^2 = x^2 -1$ has a solution, we get two solutions, $(x,\pm y)$. Thus we conclude that there should be $\frac{p-3}{2}$ distinct values of $x$ for which $(\frac{x^2-1}{p}) = 1$. Since there are two values of $x$ for which $(\frac{x^2-1}{p}) = 0$, we conclude that the remaining $p-\frac{p-3}{2} - 2$ values of $x$ evaluate $(\frac{x^2-1}{p}) $ as $-1$. Thus our sum is just $\frac{p-3}{2} - (p-\frac{p-3}{2}-2) = p-3+2-p = -1$.

user413766
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    Correct and well-written! Just a notational error: The Legendre symbol is written as $\left(\dfrac{x}{p}\right)$, not as $\dfrac{x}{p}$; the parentheses are an important part of the notation (as they disambiguate it from a plain old fraction... or, well, try to disambiguate it; they aren't great at their job). – darij grinberg Oct 04 '18 at 03:52
  • Thank you for your patience and review of this solution! I've made the suggested edits. – user413766 Oct 04 '18 at 05:15