When is the sum of consecutive squares a prime? For what integers $x$ do there exist $x$ consecutives integers, the sum of whose squares is prime?
I tried use $$1^2+2^2+...+n^2=\frac {n(n+1)(2n+1)}{6}$$

The sum of $x$ consecutive squares starting from $n$ is $$n^2+(n+1)^2+...+(n+x-1)^2=xn^2+2n\sum_{k=1}^{x-1} k +\sum_{k=1}^{x-1} k^2 $$
$$\Rightarrow n^2+(n+1)^2+...+(n+x-1)^2=xn^2+nx(x-1)+\frac{(x-1)x(2x-1)}{6}$$
If $x\equiv 1,5\ (mod\ 6)\ $ then we have $(x-1)(2x-1)\equiv0\ (mod\ 6)$
which means that $\frac{(x-1)(2x-1)}{6}\ $ is an integer so the sum becomes:
$$n^2+(n+1)^2+...+(n+x-1)^2=x(n^2+nx(x-1)+\frac{(x-1)(2x-1)}{6})$$ meaning that if $x\equiv 1,5\ (mod\ 6)\ $ the sum of the square of x consectuive integers will never be a prime.
If  $x\equiv 0\pmod6\ \text{and}\ x>6$ then $\frac {x}{6}$ will be an integer greater than 1
so we will have: $$n^2+(n+1)^2+...+(n+x-1)^2=\frac {x}{6}(6n^2+6nx(x-1)+(x-1)(2x-1))$$ and again no pirme. In the same fashion if $x\equiv2\ (mod\ 6)$ and $x>2\ $ we take $\frac{x}{2}$ as a factor, and if $x\equiv4\pmod6$ we take $\frac{x}{2}$ again as a factor and finally if $x\equiv3\pmod6\ $ and $x>3\ $ we take $\frac{x}{3}$ as a factor.
Now we need to consider tha cases in which x = 2, 3, 6:
if $x = 2$ then we have $1^2+2^2=5$ which is a prime
if $x = 3 $ then we have $2^2+3^2+4^2=29$ which is prime
and if $x = 6$ then we have $2^2+3^2+4^2+5^2+6^2+7^2 = 139$ which is again a prime.
So the answer for this problem will be 2, 3 and 6
