When is a sum of consecutive squares equal to a square? We have the sum of squares of $n$ consecutive positive integers: $$S=(a+1)^2+(a+2)^2+ ... +(a+n)^2$$ Problem was to find the smallest $n$ such, that $S=b^2$ will be square of some positive integer. I found an example for $n=11$. Now, I'm trying to prove, that if $2<n<11$ there is't any solution. So I need some help with this: If $n,a,b \in \mathbb{N}$ and $n<11$ prove that equation
$$
n\cdot a^2 + n(n+1)\cdot a + \frac{1}{6}n(n+1)(2n+1)=b^2
$$
can't be solved. Or if I'm wrong, find counterexample.
The only idea I have is: to consider the remains $\operatorname{Mod}[b^2,n]$ and $\operatorname{Mod}[\frac{1}{6}n(n+1)(2n+1),n]$. For each $2<n<11$, but it is very long.

Below is a reasonable (but not very illuminating) proof. Put $S_n(x)=\sum_{k=1}^{n} (x+k)^2$. Note that it is also true that for $2<n<11$, $S_n(x)$ is never a square modulo $n^2$, and $S_n(x)$ is also never a square modulo $900$. Perhaps this will inspire others to produce more intelligent proofs.
Here it goes :
$$
\begin{array}{ccll}
S_3(x) & \equiv & 3x^2+12x+14 \equiv 2 & {\sf mod} \  3 \\
S_4(x) & \equiv & 4x^2+20x+30 \equiv 2 & {\sf mod} \  4 \\
S_5(x) & \equiv & 5x^2+30x+55 \equiv 2 \ \text{or} \ 3 & {\sf mod} \  4 \\
S_6(x) & \equiv & 6x^2+42x+91 \equiv 3 & {\sf mod} \  4 \\
S_7(x) & \equiv & 7x^2+56x+140 \equiv 3,8,11 \ \text{or} \ 12 & {\sf mod} \  16 \\
S_8(x) & \equiv & 8x^2+72x+204 \equiv 2,5,6 \ \text{or} \ 8 & {\sf mod} \  9 \\
S_9(x) & \equiv & 9x^2+90x+285 \equiv 6 & {\sf mod} \  9 \\
S_{10}(x) & \equiv & 10x^2+110x+385 \equiv 5,10 \ \text{or} \ 20& {\sf mod} \ 25  \\
\end{array}
$$
Each time, we see that $S_n(x)$ is never a square for the given modulus.
