# solutions to equation mod a prime

I know that characterizing the solutions to an equation in a finite field is generally difficult, but I was wondering if anyone had anything to say about the equation

(ab)^2 + a^2 + b^2 = 0 mod p

I started by writing the equation as (a^2 + 1)(b^2 + 1) = 1 mod p, and also (ab + 1)^2 + (a-b)^2 = 1 mod p, but that hasn't seemed to help much. I know how to determine the number of solutions, but what I am more interested in is the (possible) implied relation between a and b.

Thanks!

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Why are you interested in an implied relation (which is at best going to be up to sign, unless you can find a canonical way to choose square roots mod $p$)? – KConrad Apr 2 '10 at 22:51
You surely know it's a rational curve. What sort of information regarding the relation between $a$ and $b$ do you want? – Robin Chapman Apr 3 '10 at 7:58

If you rewrite the equation as $b^2=-a^2/(a^2+1)$ as Will did you can continue as follows: The equation can be rewritten as $$\left(\frac{b}{a}\right)^2 = -(a^{-2}+1)$$ (excepting $a=b=0$). Putting $x=b/a$ and $y=1/a$ this gives $x^2+y^2+1=0$ or homogenising we have $x^2+y^2+z^2=0$. In any case it is a quadric and in order to find all points it is enough to just find one (one then gets all by drawing all the lines through that point and taking the second intersection point). Such a point always exists (see reference in comment) but there is no closed formula for it. Good algorithms for finding it exists however if that's what one wants.
Nice, Torsten. After I logged off I realized I had not told Sarah anything useful about which values $a$ give $( -(1 + a^2 ) | p ) = +1.$ Without your important comment about homogenising, Hardy and Wright show $1 + x^2 + y^2 = m p$ always has a solution $(x,y,m)$ with $0 < m < p.$ This is Section 6.7, Theorem 87, page 70 in my edition, title "An Introduction to the Theory of Numbers." – Will Jagy Apr 3 '10 at 16:22
For odd primes $p \; \;$: your formulation $(a^2 + 1) (b^2 + 1) \equiv 1 \pmod p$ shows that neither factor is $0 \pmod p.$ Similarly, if either $a$ or $b$ is $0 \pmod p$ so is the other. Otherwise, it is a necessary and sufficient condition (on either one of them, let us choose $a$) that Legendre symbol $( -(1 + a^2) | p ) = +1$, and then $$b^2 \equiv \frac{- a^2}{1 + a^2} \; \; .$$ So not much can be added to what you already know, and the relationship is only defined up to $\pm$ sign as commented by KConrad.
Surely you can find the solutions to the equation if ab=0. Otherwise set $a^2=1/x$ and $b^2=1/y$ to turn your equation into the equation $1+x+y=0$. Is this enough?