I came accross the following question and I can't find an easy proof of this fact : Let $p\geq 17$ be a prime number such that $p\equiv 1 \pmod 4$.

Show that for any $z\in \mathbb F_p\backslash\{0\}$, there exist $x,y\in \mathbb F_p\backslash\{0\}$ such that $$(x^2+1)(y^2+1)=xyz.$$

I tried to use combinatorial facts (such that the number of square in $\mathbb F_p$ but it is not enough) or using the existence of roots of a polynomial with one variable over $\mathbb F_p$ but it doesn't seem to be conclusive.

Any idea ?

  • $\begingroup$ What's wrong with $p=13$ adn $p=5$? is there a "bad" $z$ for these? $\endgroup$ – Hagen von Eitzen May 7 '14 at 15:12
  • $\begingroup$ For $p=5$ you can't get $z=2$ or $3$. For $p=13$ you can't get $6$ or $7$. $\endgroup$ – Robert Israel May 7 '14 at 15:19
  • $\begingroup$ ... I can prove it but only for primes $p \equiv 3 \pmod 4$ $\endgroup$ – mercio May 7 '14 at 16:07

Searching the bag of tools for a big hammer... finding this.

Fix $z$. Rewrite the equation, treating $y$ as the unknown, in the form $$ y^2-\frac{zx}{x^2+1}y+1=0. $$ This has a solution $y\in\Bbb{F}_p$, iff the discriminant $$ \Delta=\Delta(x):=\frac{z^2x^2-4(x^2+1)^2}{(x^2+1)^2} $$ is a square in $\Bbb{F}_p$. Let $\eta:\Bbb{F}_p\to\{0,\pm1\}$ be the Legendre character, i.e. $\eta(a)=+1$, if $a$ is a non-zero square, $\eta(0)=0$ and $\eta(a)=-1$, if $a$ is a non-square.

Assuming that the quartic in the numerator $p(x)=z^2x^2-4(x^2+1)^2\in\Bbb{F}_p[x]$ is not of the form constant times a square, we have the Weil bound $$ \left\vert\sum_{x\in\Bbb{F}_p}\eta(p(x))\right\vert\le (\deg p -1)\sqrt p. $$ Here $\deg p=4$, so if $p>16$ we get that $p-4\sqrt p>0$, and thus $$ \sum_{x\in\Bbb{F}_p}\eta(p(x))\ge-3\sqrt p> -p+\sqrt p. $$ Therefore there are at least $\sqrt p>4$ choices for $x$ such that $\eta(p(x))\neq-1$. Two of those choices may satisfy $x^2+1=0$, and one may be $x=0$, and these must be discarded, but at least one remains. So for $p\ge17$ we have that for some $x\in\Bbb{F}_p^*$ we have that $x^2+1\neq0$ and $\Delta(x)$ is a square. For that $x$ we have a solution $y\in\Bbb{F}_p$. Clearly the conditions on $x$ mean that $y\neq0$.

We need to deal with the possibility that for some $K,a,b,c\in\Bbb{F}_p$ we have in the ring $\Bbb{F}_p[x]$ $$ p(x)= K(ax^2+bx+c)^2. $$ Clearly then $K\neq0$. Looking at the constant terms implies that $c\neq0$. Looking at the linear terms then implie $b=0$. We are left with the possibility $$ -4x^4+(z^2-8)x^2-4=K(ax^2+c)^2. $$ Then $Ka^2=-4=Kc^2$, so $c=\pm a$. Therefore $$ p(x)=K'(x^2\pm1)^2 $$ for $K'=-4a^2$. This implies that $$ \Delta(x)=-4a^2\left(\frac{x^2\pm1}{x^2+1}\right)^2. $$ As we assume that $p\equiv1\pmod4$, we know that $\eta(-1)=1$. Thus this implies that $\eta(\Delta(x))=+1$ for all $x$ such that $x^2+1\neq0$. Again the existence of a solution follows.

  • $\begingroup$ See this question for a general description of the Weil bound. $\endgroup$ – Jyrki Lahtonen May 8 '14 at 12:02
  • $\begingroup$ Oops. Changing a single term from $-1$ to $+1$ in the character sum changes its value by two, so I'm a bit optimistic above, and the argument isn't water tight yet.. I guess it can be fixed (at least made to hold starting from some larger value of $p$). Need to commute next. $\endgroup$ – Jyrki Lahtonen May 8 '14 at 13:47

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