Techniques to prove that there is only one square in a given sequence [duplicate]

What techniques/methods can be used to prove that the sequence produced by $n\cdot (n+1)\cdot (2\cdot n+1)/6$ contains only one square ($4900$) greater than 1?

While this particular sequence is an interesting example, I'm interested in techniques that can be generalized to any sequence with a polynomial generating function.

In general, this is equivalent to asking for the solution to the Diophantine equation: $$a^2 = n\cdot (n+1)\cdot (2\cdot n+1)/6.$$

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marked as duplicate by ᴡᴏʀᴅs, Thomas, r.e.s., Claude Leibovici, user1551Sep 21 '15 at 10:30

This question was marked as an exact duplicate of an existing question.

This is not easy. The general problem of integer solutions to $f(n)=m^2$, $f$ a polynomial, is unsolved. Your particular example is an "elliptic curve", for which there is much theory. Actually, you may find solutions to your problem by searching for the "cannonball problem". – Gerry Myerson Feb 14 '13 at 6:12
@GerryMyerson, Indeed. On your advice I've found Anglin 1990 proof of this particular problem. . I will continue looking into this - perhaps I can restrict my problem domain in a way that will make it easier to land on a family of known solutions. – LBushkin Feb 14 '13 at 6:19
Shouldn't it be $a^2 = \frac{n(n+1)(2n+1)}6$ not sqrt? – Kaster Feb 14 '13 at 6:34
@Kaster - Yes. Yes it should :) – LBushkin Feb 14 '13 at 6:37
I count three squares: $0$, $1$ and $4900$. – Robert Israel Feb 14 '13 at 7:15

Note that $\gcd(n,n+1)=\gcd (n+1,2n+1) = \gcd (n,2n+1)=1$. So, $a^2=\frac{n(n+1)(2n+1)}{6} \Rightarrow \exists \ a,b,c \in \mathbb{N}$ st

Case 1: $\frac{n}{3}=r^2, \frac{n+1}{2}=s^2, 2n+1=t^2$;

So, $3r^2+2s^2=t^2$ and using the method of descent of Fermat concluded that this equation has only the trivial solution.

Case 2: $\frac{n}{2}=r^2, \frac{n+1}{3}=s^2, 2n+1=t^2$;

Analogous to the previous case.

Case 3: $\frac{n}{2}=r^2, n+1=b^2, \frac{2n+1}{3}=t^2$;

Case 4: $n=r^2, \frac{n+1}{2}=b^2, \frac{2n+1}{3}=t^2$;

Analogous to the previous case.

Case 5: $\frac{n}{6}=r^2, n+1=b^2,2n+1=t^2$;

Case 6: $n=r^2, \frac{n+1}{6}=b^2, 2n+1=t^2$;

So, our problem reduces to find all solutions of the equations $r^2+2s^2=3t^2$ and $6r^2+s^2=t^2$, but I could not solve yet...

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Apparently a proof is in the book by Mordell, Diophantine Equations (1969), page 258. – Will Jagy Apr 10 '13 at 0:56