In my current line of investigation, I am running into [many] divisibility questions like the one in the title, i.e. $$ (a+b)^2 \mid (2a^3+6a^2b+1), \qquad(\star) $$ where $a > b \ge 1$ are integers. For example, maxima calculations suggest that $(\star)$ implies $(a,b)=(4,1)$.

I really haven't the first clue how to effectively attack this problem. Any pointers and references would be greatly appreciated.

EDIT: I've now cross-posted a partial solution to this.

Kieren MacMillan
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    Just a simple observation: since $2a^3+6a^2b+1=2(a+b)^3-(6ab^2+2b^3-1)$, we must have $(a+b)^2 | (2b^3+6ab^2-1)$. – Jack D'Aurizio Oct 24 '13 at 17:21
  • Yes, I did notice that. And since $b < a$, that does provide a lower upper bound on $(a+b)^2$. But I wasn't able to get any further with the reformulation than with the original. – Kieren MacMillan Oct 24 '13 at 17:32
  • Moreover, from $(a+b)|(2a^3+6a^2 b+1)$ we have $(a+b)|(4a^3-1)$, that is equivalent to $(a+b)|(4b^3+1)$. – Jack D'Aurizio Oct 25 '13 at 09:26
  • @JackD'Aurizio: Yes, but that is much less restrictive, since that would only imply $(a,b) \in \{(4,1),(9,2),(14,11),(19,12),\dotsc\}$. – Kieren MacMillan Oct 25 '13 at 11:20

2 Answers2


$(a+b)^2|2a^3 +6a^2b+1$ iff $k(a^2+2ab+b^2) = 2a^3 +6a^2b+1$ or $-2a^3+a^2(k+6b)+a(2kb)+(kb^2+1)=0$. This is a a cubic equation so one could find $(a,b)$ by solving for $a$, but then one would still have to iterate through the $k$ and $b's$ to find an integer value of $a$ and $b$. This is also an elliptic curve, specifically $y^2 = -2a^3+a^2(k+6b)+a(2kb)+(kb^2+1)$. One could reduce this into Weiertrauss form using the cubic reduction formula to obtain something like $y^2 = a^3 + pa + q$. Anyway, elliptic curves have either finitely many or infinitely many solutions. We are searching for integer values of $a,b$ so we can apply what we know about solving for rational points on elliptic curves. If we are looking for rational points, then the rational points $E(\mathbb{Q})$ form an abelian group and is finitely generated, i.e. there is a finite number of rational points that one can use to find the rest of the points.

Solving your problem is the same as finding rational points on an Elliptic Curve and there are lots of tricks to do that, many of which can be found online. As far as I know, there is no formula for solving elliptic curves, just methods to find solutions. However, your question is even more complicated than that, because two of values in your elliptic curve are undetermined.

This link provides some examples of how to solve Elliptic Curves if your $k,b$ were determined. However, even with them determined, one would still use an algorithm to find the points. http://www.math.brown.edu/~jhs/Presentations/WyomingEllipticCurve.pdf

  • References:http://mathworld.wolfram.com/EllipticCurve.html, http://mathworld.wolfram.com/Mordell-WeilTheorem.html, http://mathworld.wolfram.com/CubicFormula.html, http://www.math.brown.edu/~jhs/Presentations/WyomingEllipticCurve.pdf – Michael Mudarri Oct 24 '13 at 17:54
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    In case it helps, the equation I'm trying to solve can be reformulated as $$ 2(2b^3+6ab^2-1) = (5b-a+1)(a+b)^2. $$ Ultimately, though, I'm interested in more general approaches. I find it fascinating that, in this case, the factor $5b-a+1$ appears to be irrelevant to the number of [integer] solutions. (Obviously, that is true in this particular case because $5b-a+1=2$ in the one known solution $(a,b)=(4,1)$, and so it cancels the factor of $2$ on the left-hand side. But how to prove that algebraically?) – Kieren MacMillan Oct 24 '13 at 18:11
  • While the original question can be reduced to finding the integral points on a parametrized family of elliptic curves (certainly not just one), it's extremely unclear that his approach is even slightly helpful for actually solving the problem. – Mike Bennett Oct 24 '13 at 18:33
  • It isn't helpful. That is what sort of what I was trying to say. Even using Elliptic Curves wont provide much insight to the solution of this problem. – Michael Mudarri Oct 24 '13 at 18:35
  • Note that in simplifying MacMillan's equation we have $(a^3-b^3) - (a^2+b^2) + 4(ab^2-a^2b) - 2(ab+1)$, and $(a^3-b^3), (a^2+b^2), (ab^2-a^2b), (ab+1)$ are linearly independent over the rationals, so we may as well be solving the equation $x-y+4z-c=0$.....edit: this may be wrong actually – Michael Mudarri Oct 24 '13 at 18:41
  • If you tried to find a good substitution for those values and try to make an elliptic curve out of it, you would be left with the huge problem of dealing with $a^3-b^3 = c^3$, for integers...I could be wrong here. – Michael Mudarri Oct 24 '13 at 18:48
  • Isn't $ab^2-a^2b=ab(b-a)$, and hence not independent of $a^3-b^3=(a-b)(a^2+ab+b^2)$? – Kieren MacMillan Oct 24 '13 at 19:04
  • Yeah, I was wrong.(I am not the most careful person here) Additionally, I think that there is a way to make an elliptic curve whose solutions are $\frac{a}{b}$ using what we have here. – Michael Mudarri Oct 24 '13 at 19:07
  • $x=\frac{a}{b}$, so $x^3 - 4x^2 + (4-\frac{1}{b})x - (1+\frac{1}{b})=0$. So we can put conditions on $b$ since we want a certain discriminant. We want a rational $x$, so this should provide conditions that need to be met when using the cubic formula or whether or not there are two nonreal, 1 real solution, or 3 real. If there were two nonreal then yeah, we would find that we only have $(4,1)$ and multiples of it. – Michael Mudarri Oct 24 '13 at 19:28
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    For $(a,b)=(4,1)$ we have $k=9$, that is a square. This strongly recalls an old and famous IMO problem, that could be attacked through Vieta jumping (http://en.wikipedia.org/wiki/Vieta_jumping). Can we say something, with the same technique, about our equation, $(k(a+b)-6a^2)(a+b)=(1-4a^3)$, since it is a quadratic equation in $b$? – Jack D'Aurizio Oct 25 '13 at 13:35
  • For sure, if $(a,b)$ is an integer solution, $(a,\frac{6a^2}{k}-2a-b)$ is a rational solution with the same $k$. – Jack D'Aurizio Oct 25 '13 at 13:41
  • By taking $c=(a+b)$, we are looking for integer solutions of $$(\heartsuit)\qquad k\, c^2 = 1-4a^3+6a^2c = 1+2a^2(3c-2a),$$ where $a < c < 2a$, i.e. $\frac{c}{2} – Jack D'Aurizio Oct 25 '13 at 15:14
  • @JackD'Aurizio: Please email me to discuss this problem further. Thanks! Kieren. – Kieren MacMillan May 08 '14 at 13:55

This is not even close to a complete solution, it will just help to reduce the number of calculations that you have to do on Maxima.

Consider $2a^3 + 6a^2b + 1 \equiv 1 \pmod{2}$ and $(a+b)^2 \equiv 1 \pmod 2$. We also have that $2a^3 + 6a^2b + 1 \equiv 0 \pmod{3}$ and $(a+b)^2 \equiv 1 \pmod{3}$. Combining these we have that $a+b = 6k+1$ for some $k\in\mathbb{Z}$.