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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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    $\begingroup$ I don't know about "elementary solutions" but Pari/GP via the command thue (but check the docs as it needs initialization) reports that the solution set is as you expect. $\endgroup$ – Felipe Voloch Sep 30 '13 at 13:02
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    $\begingroup$ Notice that the $\mathbb{Z}/2\mathbb{Z}$ symmetry $(r,s) \mapsto (-r,-s)$ comes in fact from a $\mathbb{Z}/4\mathbb{Z}$ symmetry $(r,s) \mapsto (s,-r)$. $\endgroup$ – Alberto García-Raboso Sep 30 '13 at 14:03
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    $\begingroup$ The equation is equivalent to $(r^2+2rs-s^2)^2-2(2rs)^2=1$, don't know if this relation to the Pell equation $x^2-2y^2=1$ is helpful. $\endgroup$ – Peter Mueller Sep 30 '13 at 14:14
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    $\begingroup$ @Kieren: I believe the question is more difficult than it looks at a first glance. The fact that there are non-trivial solutions makes it unlikely that some easy ad hoc argument works. One other observation I'm not sure if it helps: Your equation is $s^4h(r/s)=1$, where $h(x)$ has the cyclic group of order $4$ as Galois group, as can be seen by the fact that $x\mapsto (x-1)/(x+1)$ maps roots of $h(x)$ to roots. $\endgroup$ – Peter Mueller Sep 30 '13 at 18:51
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    $\begingroup$ I don't know if this helps but with $z=r+is \in \mathbb Z[i]$ the equation reduces to $Im(z^4(1+i))=1$. $\endgroup$ – Nick S Oct 2 '13 at 0:34
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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.

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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.

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  • $\begingroup$ Thanks! I'll look through the paper, and if it appears you're right, I'll come back and accept this as the answer. $\endgroup$ – Kieren MacMillan Sep 30 '13 at 22:46
  • $\begingroup$ I appreciate your answer, but my congruence/divisibility work (see main post) bolsters my intuition that there is a much easier proof in this special case. I'll give it a couple of weeks of poking around, and then come back to accept your answer if I've given up. Thanks again! $\endgroup$ – Kieren MacMillan Oct 3 '13 at 15:45
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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.

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    $\begingroup$ "can be shown" how? It seems unlikely that this kind of elementary manipulation can solve the problem. $\endgroup$ – Noam D. Elkies Oct 5 '13 at 3:37
  • $\begingroup$ From ($\star$), we see that for any prime $p \mid r$, we have $p^2$ dividing exactly one of the factors on the right-hand side (because they're pairwise relatively prime). This, in concert with the previous divisibility restrictions, forces $t$ out of the possible range of solutions. At least, I believe so — I'm double-checking my proof now… $\endgroup$ – Kieren MacMillan Oct 5 '13 at 13:04
  • $\begingroup$ Yes, I believe I can prove $r^2 \mid (t^2+t)$, so $t=0$. $\endgroup$ – Kieren MacMillan Oct 5 '13 at 14:32
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    $\begingroup$ What's stopping some prime factor(s) of $r$ from dividing $2t^2+2t+1$ with multiplicity $\geq 2$? $\endgroup$ – Noam D. Elkies Oct 5 '13 at 16:01
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    $\begingroup$ @Kieren: I suggest that you restrict your many updates to real progress. Also, instead of asking if or erroneously claiming that certain things work, you better first check the details for yourself. By the way, you never said why you are so much interested in this one specific equation. $\endgroup$ – Peter Mueller Oct 5 '13 at 22:10

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