Solving the quartic equation $r^4 + 4r^3s - 6r^2s^2 - 4rs^3 + s^4 = 1$

Question

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 $4/3 < r/s < 3/2$.

Answer 1

Answered on stackexchange where Kieren MacMillian linked to this MO question. Briefly: This Thue equation is equivalent to Ljunggren's equation $X^2+1 = 2Y^4$. Ljunggren proved in 1942 that the only integer solutions are $(X,Y) = (\pm 1, \pm 1)$ and $(\pm 239, \pm 13)$. [Equivalently, $(a,b,c) = (119,120,169)$ is the unique Pythagorean triple with $a-b = \pm 1$ and $c$ a perfect square.] This implies the desired result that the only solutions of $$ Q(r,s) = r^4 + 4 r^3 s - 6 r^2 s^2 - 4 r s^3 + s^4 = 1 $$ are the known small ones with $r^2+s^2 = 1$ or $13$. The implication we need is provided by the identity $Q(r,-s)^2 + Q(r,s)^2 = 2(r^2+s^2)^4$. Ljunggren's proof is a difficult application of Skolem's $p$-adic method. In the decades since then other techniques have been developed which make the solution of such an equation routine; for example this is how gp can calculate almost instantaneously

[[-2, 3], [2, -3], [0, 1], [0, -1], [3, 2], [-3, -2], [1, 0], [-1, 0]]

in response to the command

thue(thueinit(r^4+4*r^3-6*r^2-4*r+1),1)

But none of these techniques is elementary either, and as far as I know the problem of finding an elementary proof of Ljunggren's theorem remains open.

Answer 2

I just googled quartic Thue equation, and found this paper by Lettl and Petho which solves a $1$-parameter family of Thue equations where your equation is a special case. It does not look like that for your special case things could become much easier.

Answer 3

Inspired by @PeterMueller, I believe I found a proof that $r = 3$.

Because of how this equation was obtained in the first place, I can assume $s \ge 2$ is even, and $r \ge s+1$ is odd. Writing $s=2v$ and $r=2v+2t+1$, substituting, and factoring yields \begin{align} 2v(v-2t-1)(2v+2t+1)^2 &= t(t+1)(2t^2+2t+1), \end{align} at which point a reverse substitution gives \begin{align} (v-2t-1)sr^2 &= t(t+1)(2t^2+2t+1). \qquad(\star) \end{align} For any hypothetical solution with $r > 3$, we have $3s > 2r$ (shown earlier). A quick calculation then yields $t < (r-2)/4$, which can be shown to contradict ($\star$).

Does that look right?

Thanks!
Kieren.