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 $$1^2+2^2+...+n^2=\frac {n(n+1)(2n+1)}{6}$$
 A: Let our numbers be $a^2,(a+1)^2,\dots,(a+x-1)^2$. The sum of the squares is $a^2 x+2a(1+\cdots+(x-1))+(1^2+\cdots+(x-1)^2)$.
Note that $a^2 x$ is divisible by $x$, as is $2a(1+\cdots+(x-1))$. So we concentrate on the term $1^2+2^2+\cdots+x^2$. Call this number $N$. By the formula quoted in the post, we have 
$$6N=(x-1)(x)(2x-1).$$
Suppose that $x\gt 6$. Then $N$ has a factor $d\gt 1$ such that $d\mid x$. We show that $d$ is a proper divisor of $a^2+(a+1)^2+\cdots +(a+x-1)^2$. This is because 
$$a^2+(a+1)^2 +\cdots +(a+x-1)^2 \gt \left(\frac{x-1}{2}\right)^2,$$
and $\left(\frac{x-1}{2}\right)^2\gt x$ if $x\gt 6$. Thus all candidate $x$ are in the interval from $1$ to $6$. 
We can rule out $x=4$, since in that case our sum of squares is even, and clearly cannot be $2$. 
We can also rule out $x=5$, because if $x=5$ then $5$ divides $N$, and it is easy to verify that a sum of $5$ consecutive squares must be greater than $5$.
That leaves $x=1$ (no good), $x=2$, $x=3$, and $x=6$. For each of $x=2$, $x=3$, and $x=6$, we can produce examples of a sum of squares of $x$ consecutive integers which is prime. The simplest example for $x=3$ uses the consecutive integers $-1$, $0$, and $1$. The simplest example for $x=6$ uses $-2,-1,0,1,2,3$. There are also examples with all entries positive.
It is not known whether there are infinitely many solutions. 
A: The sum of the squares of the first $x$ consecutive integers, starting from $n+1$, equals
$$(n+1)^2+(n+2)^2+\ldots+(n+x)^2=\frac{(n+x)(n+x+1)(2(n+x)+1)}{6}-\frac{n(n+1)(2n+1)}{6}$$ $$=\frac{1}{6}\cdot x\cdot(6n^2+6nx+2x^2+6n+3x+1).$$
For this to be prime we must have $x\mid 6$ or 
$$6n^2+6nx+2x^2+6n+3x+1\mid6.$$
Solving the quadratic equation
$$6n^2+6nx+2x^2+6n+3x+1=c,$$
for $n$ where $c\mid 6$ yields the solutions
$$n=-\frac{1}{2}(x+1)\pm\frac{1}{2}\sqrt{3+6c-3x^2},$$
which shows that $c\geq-\tfrac{1}{2}$ and $x\leq\sqrt{1+2c}$, which reduces to $c\in\{2,3,6\}$ and $x\in\{2,3\}$ as $x=1$ is impossible. The only values that yield integral solutions for $n$ are $c=3$ and $x=2$, with solutions $n=0$ and $n=3$ corresponding to the sums of squares
$$1^2+2^2=5\qquad\text{ and }\qquad 4^2+5^2=41.$$
Otherwise $x\mid 6$. For $x=2$, $x=3$ and $x=6$ we may take $n=1$ to find
$$2^2+3^2=13,\qquad 2^2+3^2+4^2=29,\qquad 2^2+3^2+4^2+5^2+6^2+7^2=139,$$
which are all prime, and $x=1$ is still impossible. Hence the values of $x$ for which there exists $x$ consecutive integers, the sum of whose squares is prime, are $2$, $3$ and $6$.
A: 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
