I am trying to prove the following conjecture.

Conjecture. If $r > s \ge 1$ are relatively prime integers such that \begin{equation} (r-s)^4-1 \equiv 0\!\pmod{4r^2s}, \tag{1} \end{equation} then $r-s = 1$ or $2r > 3s$.

A brute-force computer search has found only solutions with $r-s=1$ and the two additional solutions $(r,s)=(10,3)$ and $(r,s)=(255,4)$.

Included below is the proof I have so far. I have attempted to extend the classic Vieta jumping technique to a cubic. My main question is: Having proved my conjecture for a cubic with three real roots, must I continue with the case where the cubic has two complex conjugate roots? Or is my claim valid, that fixing $k$ means I only have to consider the real roots?

Proof (possibly incomplete?). Evidently the congruence is satisfied when $r-s=1$. Now assume $r-s > 1$, and write $g=r-s$. Since $r$, $s$, and $g^4-1$ are all positive, (1) now implies $g^4-1 = 4kr^2s$ for an integer $k \ge 1$. Hence \begin{equation} g^4-1 = 4kr^2s = 4k(g+s)^2s = 4ks^3+8g ks^2 + 4g^2 k s, \tag{$\star$} \end{equation} which we can rewrite as the cubic equation \begin{align*} 4ks^3 + 8g ks^2 + 4g^2 ks + (1-g^4) = 0. \end{align*} Fix $k$. Dividing through by the leading coefficient $4k$ and making the substitution $s \mapsto w$ yields \begin{align} w^3 + 2g w^2 + g^2 w + \frac{1-g^4}{4k} = 0. \end{align} This cubic equation has three roots; call them $w_1, w_2, w_3$. Since $s$ is one root, fix $w_1=s$ without loss of generality. Now Vieta's formulas give \begin{align} -2g = s + w_2 + w_3, && g^2 = sw_2+ sw_3 + w_2w_3, && \frac{g^4-1}{4k} = sw_2w_3. \end{align} Combining the first two relations implies \begin{align*} 4g^2 = (-2g)^2 = (s + w_2 + w_3)^2 = s^2 + w_2^2 + w_3^2 + 2(sw_2+ sw_3 + w_2w_3) = s^2 + w_2^2 + w_3^2 + 2g^2, \end{align*} which gives \begin{equation} 2g^2-s^2 = w_2^2+w_3^2. \tag{2} \end{equation} A cubic function either has three real roots, or it has one real root and two nonreal complex conjugate roots. Since $w_1=s$ is real, $w_2$ and $w_3$ are either both real or they are nonreal complex conjugates. Since we have fixed $k$, and are seeking an integer solution with $1 < g = r-s$, we may take $w_2,w_3 \in \mathbb{R}$, where [at least] one of them is a second solution of ($\star$) for our fixed quotient $k$.

Now (2) implies $2g^2-s^2 = w_2^2+w_3^2 > 0$, yielding \begin{equation*} s^2 < 2g^2 = 2(r-s)^2. \end{equation*} Therefore \begin{equation*} \biggl(\frac{s}{r-s}\biggr)^{\!2} < 2 \qquad\implies\qquad \frac{r}{s} > \frac{1+\sqrt{2}}{\sqrt{2}} > \frac{3}{2}, \end{equation*} and hence $2r > 3s$ as claimed.

  • $\begingroup$ Without proving the $w_2,w_3 \not\in\mathbb{R}$ case, my intuition says that what I've proven here is that there is a unique integer solution $(r,s)$ for each [fixed] quotient $k$. $\endgroup$ – Kieren MacMillan Sep 7 '14 at 23:24

Take your equation $4ks^3 + 8g ks^2 + 4g^2 ks + (1-g^4) = 0$ and solve for $k$ to get $$k={\frac {{g}^{4}-1}{s (g+s)^2 }}.$$ First substitute $s=w$ in the equation and then substitue for $k$ to get (after dividing out $1-k^4$, and clearing the denominator) $${w}^{3}+2\,g{w}^{2}+{g}^{2}w-{g}^{2}s-2\,g{s}^{2}-{s}^{3}=0 $$ Since we know that $w=s$ is a root we have the courage to factor and end up with $$ \left( w-s \right) \left({w}^{2}+w(s+2\,g)+ (g+s)^2 \right) =0. $$

Evidently, since $g,s \gt 0$, the other two roots are complex.

This doesn't rule out solutions (obviously) since there are at least two. But it does mean this attempt to generalize the Vieta-jumping technique to higher than degree two does not look promising. I don't know how classic it is. I learn from Wikipedia that around 20 years ago it solved a tricky Math Olympiad problem and in following years many problems have needed that technique. But perhaps it should be better known than it is.

I wondered if one could (in search of a contradiction) set $r=s+t$ with $2t \lt s$ and perhaps further let $s=(m-1)t+u$ with $1 \le u \lt t$ and appropriate things relatively prime. So $t^4-1$ is a multiple of $4(mt+u)^2(mt+t+u).$ It seems promising to me but I got bogged down. Note that $t \gt 4m^2(m+1)$ Also $t$ is odd, less than $s/2$ and such that $t^4 \equiv 1 \bmod s.$ This means $s$ can't be a prime or twice a prime. It is not hard to get $m^4-u^4 \equiv 0 \bmod mt+u$ and other such results but I don't see how to fit them all together.

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  • $\begingroup$ Nice! But what does it say, exactly? It doesn't prove that there aren't any real/integral solutions (since there are at least the two others I've found), nor does it seem to prove $2r > 3s$, as I need to prove my conjecture. $\endgroup$ – Kieren MacMillan Sep 24 '14 at 18:06
  • $\begingroup$ Isn’t your solution for $k$ missing a factor of $4$ in the denominator? $\endgroup$ – Kieren MacMillan Apr 7 at 20:39

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