# Finding integer points on elliptic curves via divisibility conditions like $(a+b)^2 \mid (2b^3+6ab^2-1)$

Is the following conjecture correct?

Conjecture. The divisibility condition $(\alpha+\beta)^2 \mid (2\beta^3+6\alpha\beta^2-1)$ has no solutions in positive integers $1 \le \beta < \alpha < 2\beta$.

This question is related to finding integer points on a Mordell curve. A computer search outside the range indicated turned up the expected single solution $(a,b)=(4,1)$ [corresponding to the single integer point on the original Mordell curve], as well as an unexpected solution $(\alpha,\beta)=(11364,46061)$, which I can't explain. Any insights would be appreciated.

I believe a method of solution — particularly by descent — to this special case would be immediately applicable to a large class of elliptic curves.

FWIW, I've developed a partial proof which I include below. The Vieta jump implies that, for any solution $(a,b)$, there is a rational solution $(b,\tfrac{6b^2}{k}-2b-a)$ with the same $k$; in the case of the one known solution $(a,b)=(4,1)$, we do have $k=1$ in the original equation $$(5b-a+1)(a+b)^2 = 2(2b^3+6ab^2-1),$$ which makes the second Vieta root degenerate at $(1,0)$.

Proof (incomplete). The divisibility hypothesis implies $$k(\alpha+\beta)^2 = 2\beta^3+6\alpha\beta^2-1, \tag{1}$$ for an integer $k \ge 1$. Rearranging (1) and replacing $\alpha$ with the variable $\xi$ yields \begin{equation*} k\xi^2 + 2\beta(k-3\beta)\xi + (k\beta^2-2\beta^3+1) = 0.% \label{EQ: solve this} \end{equation*} One root of this equation is $\xi_1 = \alpha$. By Vieta's formulas, the other root may be written as \begin{align} \xi_2 &= \frac{2\beta(3\beta-k)}{k} - \alpha = \frac{\beta^2(k-2\beta)+1}{k \alpha}. \tag{2} \end{align}

First, assume $\xi_2$ is an integer. Since $\alpha$ is an integer, the first relation in (2) implies that the fraction must also be an integer. Hence $k \mid 6\beta^2$. But (1) implies both that $k$ is odd, and that $\gcd(k,\beta)=1$. Hence $k \mid 3$, so $k = 1$ or $3$. If $k = 3$, then the second relation in (2) implies that $\xi_2$ is positive when $\beta = 1$, and negative when $\beta > 1$. On the other hand, the first relation with $k=3$ gives $\xi_2 = 2\beta(\beta-1) - \alpha$. When $\beta = 1$, this implies $0 < \xi_2 = 2\beta(\beta-1) - \alpha = -\alpha$, contradicting $\alpha > 0$. When $\beta > 1$, we have $0 > \xi_2 = 2\beta(\beta-1) - \alpha$. Hence $\alpha > 2\beta(\beta-1)$, contradicting $\alpha < 2\beta$. END OF PARTIAL PROOF

Note that the unexpected solution $(\alpha,\beta)=(11364,46061)$ implies \begin{equation*} k = \frac{2\beta^3+6\alpha\beta^2-1}{(\alpha+\beta)^2} = \frac{340107729770625}{3297630625} = 3 \cdot 31 \cdot 1109, \end{equation*} and then we have \begin{equation*} \frac{2\beta(3\beta-k)}{k} - \alpha = \frac{\beta^2(k-2\beta)+1}{k \alpha} = \frac{685486248}{34379} \end{equation*} not an integer.

• Why do you think the conjecture is true? – GH from MO Aug 23 '14 at 18:32
• Maxima brute-force calculations, mostly. A different form of it might be true: If $\alpha > \beta$ satisfies the hypothesis, then the quotient is $1$ or $\alpha > 2(\beta+1)$. I'd also be happy to have a counterexample found. – Kieren MacMillan Aug 23 '14 at 18:36
• As I said in the MSE thread, the exact equation I'm trying to solve can be written as $(5b-a+1)(a+b)^2=2(2b^3+6ab^2-1)$, and I happen to know, based on how I derived this form, that the only solution is $(a,b)=(4,1)$. But I'm interested in figuring out a general method to attack this class of divisibility problems. – Kieren MacMillan Aug 23 '14 at 18:39
• See also the meta thread: meta.mathoverflow.net/questions/1865/… – Todd Trimble Aug 23 '14 at 21:36
• Kieren, I hope you don't mind the edit. I just wanted to make sure the post was in the form of a question. – Todd Trimble Aug 24 '14 at 2:51

Here's a reformulation of the conjecture, and a heuristic which suggests that the conjecture is false. Note that $a+b$ must be odd. Put $\ell = a+b$ (and $\ell$ is odd) and $m=a-b$ which is also odd. The condition that $b<a<2b$ now translates to $0<m < \ell/3$. The divisibility condition of the question translates nicely to the condition $\ell^2$ divides $m^3-2$.

So the question really is can $m^3-2$ have a square factor of size exceeding $(3m)^2$? Here's a heuristic that indicates why this should be possible. Given $\ell$ odd, the number of solutions $\pmod {\ell^2}$ to $x^3-2\equiv 0 \pmod{\ell^2}$ is given by a multiplicative function $f(\ell)$. Here if $p\ge 5$ then $f(p^k)$ is simply the number of solutions to $x^3\equiv 2 \pmod p$ which is $0$, $1$ or $3$ (with asymptotic densities as in Chebotarev). For $p=3$ one should work a tiny bit harder. Then the question is whether one of these $f(\ell)$ solutions $\pmod{\ell^2}$ happens to lie below $\ell/3$. We may expect that probability to be about $f(\ell)/(3\ell)$. Thus the expected number of $\ell$ below $x$ for which $\ell^2$ divides $m^3-2$ (with $m< \ell/3$) should be about $$\sum_{\ell < x; \ell \text{ odd}} f(\ell)/(3\ell) \sim C \log x,$$ for some positive constant $C$ -- with a little calculation, using the Dedekind zeta function one should be able to calculate $C$. This suggests that there are infinitely many counterexamples.

I didn't find any with $m$ below $10000$ (I searched for large square factors of $m^3-2$), but my guess would be that there is a reasonable chance of finding counterexamples with $m$ in the millions, and a very good chance if $m$ is in the billions.

Added: A relevant paper for such problems is Granville's paper in IMRN.
Granville shows that on the abc conjecture the largest square factor of any cubic polynomial (without repeated roots) $f(m)$ is at most $m^{2+o(1)}$. This generalizes the observation in GH from MO's answer (the general case requires more work than the easy $x^3-2$ example). Granville also gives reasons to believe that this exponent is best possible, and in particular shows that such large square factors are attained for cubic polynomials having a rational root -- this uses the solutions to Pell's equation. Of course, Granville's construction doesn't apply for $x^3-2$ which is irreducible, but the results may be suggestive. Also note that the Pell type constructions produce a logarithmic number of examples with large square factors, which is again suggestive.

• I made your heuristic a bit more specific, and performed more extensive experiments. See my added section – GH from MO Aug 25 '14 at 14:42
• @GHfromMO: That's very nice! If we think of a Poisson process, then maybe one expects to find no counterexample around size $10^{10}$ with probability $1/e$, and finding no counterexample of size $10^{20}$ with probability $1/e^2$ etc. Also, I'm not completely sure of the numerical value you give for $C$ -- maybe a typo somewhere? (If $C$ is really as big as $0.4$ wouldn't you expect counterexamples sooner?) – Lucia Aug 25 '14 at 15:29
• @GHfromMO: Your values seem to give $C$ more like $0.04$ rather than $0.4$? (I mean the $C$ in the $C \log x$ of my answer.) This is why I thought there may be a typo there. – Lucia Aug 25 '14 at 15:47
• I have now found a pair with $m/\ell\approx 0.766$! See my added section. – GH from MO Aug 25 '14 at 20:40
• We are now at $m/\ell\approx 0.485$ which is smaller than all previous ratios, including the one for $(\ell,m)=(5,3)$. – GH from MO Aug 26 '14 at 1:03

This is a supplement to Lucia's answer. His/her heuristic analysis suggests that there are infinitely many odd $m$'s such that the largest square dividing $m^3-2$ exceeds $9m^2$, contradicting the conjecture.

On the other hand, the abc conjecture implies that the largest square dividing $m^3-2$ is $\ll_\epsilon m^{2+\epsilon}$ for any $\epsilon>0$, which indicates that a slightly weaker version of the conjecture (i.e. one with a more restrictive constraint on the variables) is probably true.

Some calculations with SAGE show that for $m<10^7$ there are only two instances where the largest square dividing $m^3-2$ exceeds $m^2$, namely $m=3$ and $m=100$. (Noam Elkies extended the range to $m<\ell<10^8$ with a more efficient gp-pari code.) I found five further values with a square divisor exceeding $m^2/2$, and these are $\{1244,11317,296428,714417,722428\}$.

Added. The constant in Lucia's heuristic analysis (with $m$ restricted to odd numbers) equals $$\frac{1}{6}\frac{1}{2}\frac{2}{3}\prod_{\substack{p>3\\f(p)=0}}\left(1-\frac{1}{p^3}\right)\prod_{\substack{p>3\\f(p)=1}}\left(1-\frac{1}{p^2}\right)\prod_{\substack{p>3\\f(p)=3}}\left(1-\frac{3}{p^2}+\frac{2}{p^3}\right)$$ times the residue of the Dedekind zeta function of $\mathbb{Q}(\sqrt[3]2)$. The initial factor $1/6$ accounts for Lucia's denominator $3$ and the fact that we only look for odd $m$'s. The other factors come from comparing the Euler factors of $\sum_{\ell=1}^\infty f(\ell)\ell^{-s}$ to those of the Dedekind zeta function. My calculation in SAGE shows that $C\approx 0.0423$, which suggests that the average ratio between the consecutive $\ell$'s yielding a counterexample $(\ell,m)$ is about $1.867\times 10^{10}$.

Summary of ongoing calculation. Counterexamples to the OP's conjecture correspond bijectively to pairs of odd numbers $(\ell,m)$ such that $\ell^2\mid m^3-2$ and $m/\ell<1/3$. I extended Noam Elkies' gp-pari calculation to the range $\ell<10^{12}$ and am checking higher ranges on several machines. The best pairs I found so far (allowing $m$ even) are $$(\ell,m)=(10444012561, 5062142741)\quad\text{with ratio}\quad m/\ell\approx 0.485$$ $$(\ell,m)=(22713683537, 7950843140)\quad\text{with ratio}\quad m/\ell\approx 0.350$$ In particular, these beat the basic ratio $m/\ell=0.600$ that corresponds to $(\ell,m)=(5,3)$.

• Note that the heuristic I gave also suggests that apart from finitely many exceptions, the largest square factor of $m^3-2$ is at most $(m\log m)^2$, say. So this is certainly in a very delicate range! – Lucia Aug 24 '14 at 20:21
• It's faster to look for $\ell^2 \mid m^3 - 2$ via the factorization of $\ell$ rather than $m^3-2$. A 30 minute calulation in gp (using polrootspadic) finds that there are no examples of $m<\ell$ with $\ell$ odd and $0 < \ell < 10^8$ besides the known $(\ell,m) = (5,3)$ and $(127,100)$. – Noam D. Elkies Aug 25 '14 at 0:59
• OK, gp code follows, though I'm afraid the line breaks and indentations will be lost $-$ or is there a way to retain such formatting in comments?  L = 10^8; { forstep(l=3,L,2, if(l%10^6==1,print("<",l-1,">")); if(l%3, F = factor(l); n = #F[,1]; v = vector(n,i,polrootspadic(x^3-2,F[i,1],2*F[i,2])); for(i=1,n, v[i] = lift(v[i]) * Mod(1,F[i,1]^(2*F[i,2]))); forvec(r=vector(n,i,[1,#v[i]]), m = Mod(0,1); for(i=1,n, m=chinese(m,v[i][r[i]])); m = lift(m); if(m<l,print([l,m])) ) ) ) } – Noam D. Elkies Aug 25 '14 at 2:46
• 0.350... You're getting close! But you should consider writing explicitly in your answer, near the numerical examples, that an example with $m/l < 0.333$ would be a counter-example to the original OP's conjecture. Right now a new reader has to read carefully Lucia's answer or the comment on yours to find this information. – Joël Aug 26 '14 at 12:59
• I have checked all $\ell<10^{12}$ by distributing the task between many machines. This is quite a big range. I will continue the search for another week, month, or year (depending when I get tired), and I will post a summary then. At the moment all I can say is that the data is in good agreement with Lucia's heuristic. – GH from MO Sep 6 '14 at 4:24

I think that you can, except for your one case of $(b,j)=(1,2)$, change the bounds to $300+1 \lt b \lt a \lt 2b-300.$ The number $300$ could easily be made as large as you please (unless a counter-example to your conjecture turned up) because each $j$ only allows a few possible pairs $a=2b+j$ and these can be checked. That does not prove anything but the method is simple and might interest you.

Pick a particular $j$ (and you have assumed $j \le 2$), set $a=2b+j$ and consider first the weaker condition that $a+b \mid 2b^3+6ab^2-1$ which becomes $3b+j \mid 14b^3+6b^2j-1$. Any common divisor of those two must also divide $14b^2(3b+j)-3(14b^3+6b^2j-1)=-4b^2j+3.$ Continuing in this fashion we see that any common divisor must divide $27-4j^3.$ So

• find all the divisors of $27-4j^3$ of the form $3k+j$

• For each set $b=k,a=2k-j$ and, provided that $b \lt a$ check if indeed $3b+j \mid 14b^3+6b^2j-1$.

• On the rare occasions that that happens check the actual desired condition that $(3b+j)^2 \mid 14b^3+6b^2j-1$.

For $-30 \le j \le 2$ the pairs $[b,j]$ which survive until the last test are

$[2671, -28], [5865, -28], [451, -26], [3254, -23], [625, -22],$$[843, -22], [322, -19], [179, -18], [602, -17], [18, -13],$$ [76, -13], [592, -13], [11, -8], [141, -8], [12, -5], [1, 2]$

But only $[1,2]$ survives the last step.

I checked as far as $j=-300.$