# A congruence conjecture regarding $(r-s)^4-1 \equiv 0\!\pmod{4r^2s}$

Is the following conjecture true?

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

A brute-force computer search has so far found only solutions with $r-s=1$ and the two additional solutions $(r,s)=(10,3)$ and $(r,s)=(255,4)$. [n.b. Noam Elkies confirmed the conjecture up to $r = 3 \cdot 2^{22} > 1.25 \cdot 10^7$ using gp.]

I posted a partial proof on MSE, but got no help despite several upvotes and a bounty offer (which has since expired).

The motivation for the proof is the application of Vieta jumping to equations greater than the second degree, primarily as a method of attacking Thue equations. So although any proof of the conjecture would be nice, a completion of my partial proof would be preferred.

• I am not sure, if I have got the modulus notation right here. Would you be so kind and calculate an example, say for $r=19$ and $s=17$ ? I am quite stuck at the point that the left side gives 15, and the modulus is built with 24548. So shouldn't be the result 15 instead of 0 ? – Johannes Trost Sep 24 '14 at 16:47
• Well, I can only calculate an example that works, so $(r,s)=(19,17)$ is out, I think. But taking the first stated ”additional solution” $(r,s)=(10,3)$, we have $$(r-s)^4-1=7^4-1=2400=(4 \cdot 10^2 \cdot 3) \times 2,$$ so the quotient is $2$. – Kieren MacMillan Sep 24 '14 at 17:02
• Got it. I've misread the conjecture. Apologies. – Johannes Trost Sep 25 '14 at 7:38
• How big was your computer search? What were the ranges for $r, s$? – Mayank Pandey Oct 10 '14 at 2:44
• Noam Elkies confirmed the conjecture up to $r = 3 \cdot 2^{22} > 1.25 \cdot 10^7$. – Kieren MacMillan Oct 10 '14 at 2:55

You are attempting to generalize to cubics, a form of infinite descent sometimes used with the aid of quadratic polynomials. In short (see the link for a better description) one assumes a certain positive integer pair is the minimal solution of some problem, produces a quadratic polynomial they solve, adjusts it to get another, and that the new polynomial gives a smaller valid solution.

As I commented on MSE (before giving a possible approach which does not merit repeating): Your partial proof uses a cubic with a known integer root, $w_1,$ and then discusses the case that the other two roots are real (perhaps not integers). However for this problem they will never be real, so that entire line of argument does not seem promising as a vehicle for generalizing the technique to polynomials of degree $3$ and higher.

If that is a strong motivation, then perhaps another problem would be better.

The link also shows that around 20 years ago this technique solved a tricky Math Olympiad problem and, in following years, a number of the set olympiad problems have had nice solutions using it. I'm not sure how successful it has been with problems not designed to utilize it. But perhaps it should be better known than it is.

• One of the reasons that I feel Vieta-jumping (or an extension thereof) might be applicable here is that the two known “additional solutions” exhibit a Vieta-jump-like descent/association, i.e., it may be that $(10,3) \mapsto (3,2)$ and $(255,4) \mapsto (4,3)$, though I haven't shown such a connection explicitly. That sort of $(a,b) \mapsto (b,b')$ descent/association is exactly what [quadratic] Vieta-jumping does so well. – Kieren MacMillan Oct 10 '14 at 2:07
• That is intriguing. Of course small numbers facilitate coincidences. I've had that break my heart more than once. In the quadratic case can you also climb and get bigger examples? If so, can you do that here (perhaps losing the $2r \gt 3s$ condition?) – Aaron Meyerowitz Oct 10 '14 at 2:22
• The Strong Law of Small Numbers is my kryptonite. – Kieren MacMillan Oct 10 '14 at 2:46
• Of course one always has a solution $(s,s-1),$ so it is not such a coincidence. – Aaron Meyerowitz Oct 10 '14 at 5:31
• Yes, that’s why I say "it may be that…". – Kieren MacMillan Oct 10 '14 at 13:25