Techniques to prove that there is only one square in a given sequence 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:
$$
a^2 = n\cdot (n+1)\cdot (2\cdot n+1)/6.
$$

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...
