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I'm looking to find integer solutions for large positive $y$ values (say over 1000) to the following equation:

$210y^2=(x)(x+1)(2x+1)$

What I know so far:

Integer solutions include (0,0) and (7,2)

Rational Solutions include (2/3, 1/9), (3/4, 1/8), (1/4, 1/49), (1/5, 1/25) and (125/722, 495/13718).

I've found these by taking any two points I already know (found from guess and check to start), finding the equation of the line between them, and then using Wolfram Alpha to find the third point for me (reasoning that any cubic equation must have three solutions, and if two are rational, so too must be the third, as irrational roots come in pairs).

Also, I know that $x$, $x+1$, and $2x+1$ are all coprime, and must somehow be composed of 2, 3, 5, 7, and square numbers, but this isn't leading me very far.

I have looked in to Vieta jumping/root flipping, setting $x*(x+1)*(2x+1)/210=k$ and trying to show that k must be a square number, but I've failed to do that and anyways, I don't see how that would lead me to a value of x or k.

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  • $\begingroup$ You have rediscovered the "chord and tangent" process. Well done! $\endgroup$ – Gerry Myerson Jun 18 '16 at 3:10
  • $\begingroup$ The similar-looking equation, $6y^2=x(x+1)(2x+1)$, is very famous, as it asks for a pyramid of cannonballs to contain a square number of cannonballs. I'm sure you'll find many references to it on the web, and maybe they will give you some idea of the kinds of methods used to deal with such equations. $\endgroup$ – Gerry Myerson Jun 18 '16 at 3:15
  • $\begingroup$ @GerryMyerson: Yes, he/she is essentially looking for a square pyramidal number $P_n=\frac{1}{6}n(n+1)(2n+1)$ such that $P_n = 35y^2$. (The cannonball problem has $P_n=y^2$ and has well-known solution $P_{24}$.) Robert Israel showed that $P_{840}$ is a "large" solution to dora's problem. Makes me wonder for what $d$ does $P_n = dy^2$ have interestingly large solutions. $\endgroup$ – Tito Piezas III Jun 18 '16 at 5:25
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This being an elliptic curve, Siegel's theorem says there are only finitely many integer points. But you did miss some: $(-1,0)$ and $(840, \pm 2378)$.

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