Is the following statement true? If yes, how can find the solutions?
The equation $$3x^2+8xy+7y^2\equiv-1\pmod p$$ has an integral solution for every prime $p>5$.
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2$\begingroup$ Call $q$ your quadratic form and $M$ the quadratic module $(\mathbf Z^2, q)$. Then $M\otimes \mathbf F_p$ is non-degenerate for $p>5$ (and $p=3$), and non-degenerate quadratic planes are universal, so solutions exist. Finding them doesn't seem to be as easy. $\endgroup$– few_repsCommented Dec 16, 2015 at 9:30
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$\begingroup$ Thanks for your answer. Where can I find the terminology and the universality of the non-degenerate quadratic planes? $\endgroup$– M. Farrokhi D. G.Commented Dec 16, 2015 at 10:19
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$\begingroup$ There is parametrization of all solutions if p>1 and -5 is square residue modulo p, are interested in this? $\endgroup$– joroCommented Dec 16, 2015 at 10:35
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5$\begingroup$ This curve has $p \pm 1$ points depending on what happens at infinity, as long as it's smooth (i.e. $p \ne 2,5$). There are many simple proofs of this or one can appeal to the Weil bounds. To find a solution, pick $x$ at random and solve for $y$, there is a 50% chance that a given $x$ succeeds. To solve for $y$ you can use the quadratic formula. To take square roots modulo $p$ using one of the known algorithms (google will help). $\endgroup$– Felipe VolochCommented Dec 16, 2015 at 10:39
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2 Answers
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Call $q$ your quadratic form, and $M$ the $\mathbf Z$-quadratic space $(\mathbf Z^2,q)$. It's discriminant is $-20$. Now let $p$ be a prime. Then $V_p:=M\otimes \mathbf F_p$ is a $2$-dimensional quadratic space over $\mathbf F_p$, whose discriminant is the image of $-20$ in $\mathbf F_p$. Thus this quadratic space is non-degenerate iff $p$ does not divide $20$.
It's a classical theorem that non-degenerate planes over a finite field $\mathbf F_q$ with $q$ odd are universal (i.e. represent all elements of $\mathbf F_q^{\times}$. Thus you can find solutions to your equation when $p=3$ or $p>5$.
How to find one such solution ? Fix a prime $p>5$. Then in $\mathbf F_p$, we can write $$q(x)=3(x+\frac 43 y)^2+\frac 53 y^2\ \ \ .$$ After a change of variables, the question becomes : how to find a solution of the equation $$x^2+5y^2=-3$$ in $\mathbf F_p$ ? There are obvious solutions when $-2$ is a square in $\mathbf F_p$, when $-3$ is a square in $\mathbf F_p$, when $-5$ is a square in $\mathbf F_p$, when $-15$ is a square in $\mathbf F_p$. Thus in these cases, you can easily cook a solution. In the remaining cases, $3$ is a square, $5$ is a square, $-1$ is not a square, and a new change of variables shows that is is equivalent to solve the equation $$x^2+y^2=-1$$ in $\mathbf F_p$.
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Working in the finite field $\mathbb{F}_p$ and applying a linear change of variables, the equation can be written as $$a_1x_1^2+a_2x_2^2=1$$ with some nonzero coefficients $a_1,a_2\in\mathbb{F}_p^\times$. As Felipe Voloch said, this equation has $p+1$ or $p-1$ solutions depending on whether $-a_1a_2$ is a square in $\mathbb{F}_p^\times$ or not. For a proof of this fact, in the more general setting of $k$ variables, see my response here.
answered Dec 16, 2015 at 17:06