I'm working on solving the quartic Diophantine equation in the title. Calculations in maxima imply that the only integer solutions are \begin{equation} (r,s) \in \{(-3, -2), (-2, 3), (-1, 0), (0, -1), (0, 1), (1, 0), (2, -3), (3, 2)\}. \end{equation} Evidently, the set above are all solutions, and furthermore if $(r,s)$ is a solution then so is $(-r,-s)$; hence we only need to prove that there are no solutions with $r > 3$. I have factored the equation as both \begin{align} 4r^2s(3s-2r) &= (r-s)^4-1 = (r-s-1)(r-s+1)\bigl((r-s)^2+1\bigr) \end{align} and \begin{align} 4rs^2(3r+2s) &= (r+s)^4-1 = (r+s-1)(r+s+1)\bigl((r+s)^2+1\bigr), \end{align} but don't know where to go from that point. I am hoping there is an elementary solution, even if it's not particularly "simple".

Any help would be appreciated.

Thanks,

Kieren.

EDIT: Note that for all known solutions, $\lvert r + s\rvert = 1$ or $\lvert r - s\rvert = 1$.

EDIT: In a comment, Peter M. pointed out that this can be written as the Pell equation $$ (r^2+2rs-s^2)^2-2(2rs)^2=1. $$ Curiously — and perhaps not coincidentally — the fundamental solution to that Pell equation is $(3,2)$, which is also the largest [conjectured] positive integer solution. As the fundamental solution in this case is $(r^2+2rs-s^2,2rs)$, whereas the largest integer solution is $(r,s)$, perhaps there's a way of using that to force some sort of descent or contradiction?

EDIT: Adding $4r^4$ and $4s^4$ to both sides of the equation and factoring yields, respectively \begin{align*} (r-s)^2(r+s)(r+5s) &= (2s^2-2s+1)(2s^2+2s+1) \end{align*} and \begin{align*} (r-s)(r+s)^2(5r-s) &= (2r^2-2r+1)(2r^2+2r+1) \end{align*} Note that, in each case, the two factors on the right-hand side are relatively prime (because they're odd, and evidently $\gcd(r,s)=1$). So far, this is the most interesting factorization I've found.

EDIT: Considering the equation modulo $r-s$ and modulo $r+s$, one can (I believe) prove that if a prime $p \mid (r-s)(r+s)$, then $p \equiv 1\!\pmod{4}$.

EDIT: Still holding out for an elementary proof. In addition to the restriction $$p \mid (r-s)(r+s) \implies p \equiv 1\!\pmod{4},$$ I've found the following list of divisibility restrictions: \begin{align} r &\mid (s-1)(s+1)(s^2+1) \\ s &\mid (r-1)(r+1)(r^2+1) \\ (r-s) &\mid (4r^4+1) \\ (r+s) &\mid (4s^4+1) \\ (r+s)^2 &\mid (4r^4+1) \\ (r-s)^2 &\mid (4s^4+1) \\ (r-s-1) &\mid 4(s-2)s(s+1)^2 \\ (r-s+1) &\mid 4(s-1)^2s(s+2) \\ (r+s-1) &\mid 4(s-3)(s-1)s^2 \\ (r+s+1) &\mid 4s^2(s+1)(s+3), \end{align} as well as a host of other [less immediately compelling] restrictions. Based on this, I'm hoping to prove that one of $r-s$ or $r+s$ must be $\pm 1$; bonus if I can show that the other divides $5$.

EDIT: I can show that $4s > 3r$. Calculations in maxima suggest that no numbers $r,s$ with $4 \le r \le 13000$ and $r > s \ge 1$ and $r$ odd and $s$ even and $r-s>1$ and $4s>3r$ also satisfy the six divisibility requirements \begin{align} r &\mid (s-1)(s+1)(s^2+1) \\ s &\mid (r-1)(r+1)(r^2+1) \\ (r-s-1) &\mid 4(s-2)s(s+1)^2 \\ (r-s+1) &\mid 4(s-1)^2s(s+2) \\ (r+s-1) &\mid 4(s-3)(s-1)s^2 \\ (r+s+1) &\mid 4s^2(s+1)(s+3). \end{align} Note that I didn't even need all of the congruences in the previous list. Next I'll run $r$ up to $10^6$ or so. Hopefully, though, I can obtain an algebraic proof of all of this!

EDIT: So far, the best bounds I can prove are $4/3 < r/s < 3/2$.

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