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.
 A: 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.
A: In general, there are $n$ consecutive squares that add to a square if and only if the following Diophantine equation has solutions:
$$x^2-ny^2=\frac{n(n-1)(n+1)}{3}$$
with the added parity condition that $y$ and $n$ are different parities. The value $y$ is just the sum of the first and last element of the sequence, so $$\frac{y-n+1}{2}$$ is the beginning of the sequence.
When $n$ is not  square, then the nice thing about Pell-like equations is that if there is one solution with the above parity condition, there are infinitely many. In particular, then there is always a sequence of consecutive positive numbers if there is any consecutive sequence of integers.
If $n$ is a perfect square, then there is a solution if and only if $(n,6)=1$. It is not always the case that you can find consecutive squares of positive numbers in this case. But we can get a solution $m,m+1,\dots,m+n-1$ with $m=\frac{(n+1)(n-25)}{48}$ when $n$ a perfect square, which gives positive solutions except when $n=1$ and $n=25$.  But $n=1$ obviously has positive solutions, so the only problem case is $n=25$. There is no positive solution when $n=25$, only a non-negative solution: $0^2+1^2\dots+24^2$ is a square.
When $n$ is not a perfect square, this condition gets harder, but the Pell-like nature of the equation gives us a way to find some solutions. For exmaple, $u^2-nv^2=n(n-1)$ has a trivial solution, so if $n+1$ is divisible by $3$ then we can find a solution to the above equation of $w^2-nz^2 = \frac{n+1}{3}$ yields a solution to the above, and it turns out you can easily show in that case that the solution satisfies the parity condition. In particular, if $n=3w^2-1$, then $(w,z)=(w,0)$ is a solution, so we know we have infinitely many non-square values of $n$, and, in particular, $n=11$ when $w=2$.
Necessary conditions when $n$  is not a square:
A. Factor $n=a^{2}b$, with $b$ square-free, then:


*

*If $2 \mid a$, then $2 \mid b$

*If $3 \mid a$, then $3 \mid b$

*$3$ is a square $\pmod{b}$. (Alternatively, the only primes that divide
 $b$ are $2$,  $3$, and primes of the form $12k \pm 1$.)

*If $3 \mid b$, then $b \equiv 6 \pmod{9}$


B. 


*

*If $3 \mid n+1$, then $\frac{n+1}{3}$ is the sum of two perfect squares 

*If $3 \nmid n+1$, then $n+1$ is the sum of two perfect squares


These can be used to exclude $n=3,...,10$:


*

*$3$ violates $A.4.$

*$4$ is a square and $(4,6)\neq 1$

*$5$ violates $A.3.$. 

*$6$ violates $B.2$.

*$7$ violates $A.3.$ 

*$8$ violates $B.1.$

*$9$ is a square not relatively prime to $6$. 

*$10$ violates $B.2.$


See here for some more details and some ways to explicitly construct such $n$.
I don't have sufficient conditions.  The smalles number that satisfies the above conditions but for which there is no solution is $n=842$.
It is easier to come up with necessary and sufficient conditions for the question of whether there exists an arithmetic progression of length $n$ such that the sum of the squares is a square. Indeed, if $y$ and $n$ are the same parity in the above equation, that amounts to a sequence of $n$ consecutive odd numbers whose squares add up to a square.
A: There is a solution for n=2, namely:
n=2; a=2,b=5. 
since 3 squared + 4 squared = 5 squared, 
proof: 9+16=25.
However I noticed n=2 was excluded in the equality stating n less than 11. It is not clear to me why n=2 is excluded. 
