I am working on a Diophantine equation in integers $x$ and $y$. The equation has been solved, so I already know the solutions (there are four) — I am trying to find a more elementary solution.
Through various assignments and [rational] transformations, I’ve derived a number of polynomials in $x$, $y$, and six additional integer parameters ($p$, $q$, $r$, $s$, $t$, and $k$). I’ve found many polynomials that equal zero, in terms of three or more parameters: $krs−kpr+r+q−k=0$ and $kr^2s − r^2 − qr − kr − kq + k^2 = 0$ and so on. I’m trying to use these polynomials together in order to force constraints on the original numbers $x$ and $y$, with the intention of solving the original equation.
My question is this: If I know that $p$ divides $6r+q+3k$ and $krs-5r-4k$ and $k^2+kq-5r^2$ and $r^2s+2r+2k$ and any number of other “polynomial multiples of $p$”, but I don’t necessarily know a simple expression for the quotient — and I certainly “can’t know” [a priori] the exact integer quotient, since that’s what I’m trying to prove! — are there methods of combining these “multiples of $p$” in order to prove (e.g.) that $6r+q+3k=krs-5r-4k$ [which would lead to the known solution the original equation]?
Is there, for example, a “Chinese Remainder Theorem” for polynomials, or some similar way of reducing all these equations to hard numbers?
Obviously, if I could find a “zero polynomial” in a single variable, that would be sufficient to finish the proof. Unfortunately, I’ve been able to find “zeros” in any number of three-variable combinations (e.g., $y$ and $r$ and $k$, or $p$ and $s$ and $k$, or $y$ and $p$ and $s$), but am finding it essentially impossible to reduce that to even two variables, much less a single variable.
What method(s) should I be looking into?
Assuming $p > 3$, $6 r + q + 3 k \equiv 0 \pmod p$ is equivalent to $r \equiv -q/6 - k/2 \mod p$, and you can substitute that in to $krs−5r−4k \equiv 0 \pmod p$ or any other polynomial equation mod $p$ to get a congruence that doesn't mention $r$. But you're not going to get any equation that is not mod $p$: any set of polynomial equations mod $p$ that has a solution will have infinitely many solutions. So I don't see how you could possibly conclude some expression $= 0$, rather than just $\equiv 0 \pmod p$.