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 $p>5$.



A quadratic Diophantine equation
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 $\mathbf F\_p$. 


