Solving the quartic equation $r^4 + 4r^3s - 6r^2s^2 - 4rs^3 + s^4 = 1$ 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 $4/3 < r/s < 3/2$.
 A: 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.
A: 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.
A: 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.
