, 2009

Solution:
Obviously, $p_{13} \neq 2$, because sum of squares of 12 prime numbers is greater or equal to $12 \times 2^{2}=48$. Thus, $p_{13}$ is odd number and $p_{13} \geq 7$.
We have that $n^{2} \equiv 1 \pmod{8}$, when $n$ is odd. Let $k$ be the number of prime numbers equal to $2$. Looking at equation modulo $8$ we get:
$$4k + 12 - k \equiv 1 \pmod{8}$$
So, $k \equiv 7 \pmod{8}$ and because $k \leq 12$ we get $k=7$. Therefore, $p_{1}=p_{2}=\ldots=p_{7}=2$. Furthermore, we are looking for solutions of equations:
$$28 + p_{8}^{2} + p_{9}^{2} + p_{10}^{2} + p_{11}^{2} + p_{12}^{2} = p_{13}^{2}$$
where $p_{8}, p_{9}, \ldots, p_{13}$ are odd prime numbers and one of them is equal to $p_{9}+4$.
Now, we know that when $n$ is not divisible by $3$, $n^{2} \equiv 1 \pmod{3}$. Let $s$ be number of prime numbers equal to $3$. Looking at equation modulo $3$ we get:
$$28 + 5 - s \equiv 1 \pmod{3}$$
Thus, $s \equiv 2 \pmod{3}$ and because $s \leq 5$, $s$ is either $2$ or $5$. We will consider both cases.

i. When $s=2$, we get $p_{8}=p_{9}=3$. Thus, we are looking for prime numbers $p_{10} \leq p_{11} \leq p_{12} \leq p_{13}$ greater than $3$ and at least one of them is $7$ (certainly $p_{13} \neq 7$), that satisfy
$$46 + p_{10}^{2} + p_{11}^{2} + p_{12}^{2} = p_{13}^{2}$$
We know that $n^{2} \equiv 1 \pmod{5}$ or $n^{2} \equiv 4 \pmod{5}$ when $n$ is not divisible by $5$. It is not possible that $p_{10}=p_{11}=5$, because in that case $p_{12}$ must be equal to $7$ and the left-hand side would be divisible by $5$, which contradicts the fact that $p_{13} \geq 7$. So, we proved that $p_{10}=5$ or $p_{10}=7$.

If $p_{10}=5$ then $p_{11}=7$ because $p_{11}$ is the least of remaining prime numbers. Thus, we are looking for solutions of equation
$$120 = p_{13}^{2} - p_{12}^{2}$$
in prime numbers. Now, from
$$2^{3} \cdot 3 \cdot 5 = (p_{13} - p_{12})(p_{13} + p_{12})$$
that desired solutions are $p_{12}=7, p_{13}=13$; $p_{12}=13, p_{13}=17$; $p_{12}=29, p_{13}=31$.

If $p_{10}=7$ we are solving equation:
$$95 + p_{11}^{2} + p_{12}^{2} = p_{13}^{2}$$
in prime numbers greater than $5$. But left side can give residues $0$ or $3$ modulo $5$, while right side can give only $1$ or $4$ modulo $5$. So, in this case we do not have solution.

ii. When $s=5$ we get equation:
$$28 + 45 = 73 = p_{13}^{2}$$
but $73$ is not square or integer and we do not have solution in this case.

Finally, only solutions are:
$\{(2,2,2,2,2,2,2,3,3,5,7,7,13), (2,2,2,2,2,2,2,3,3,5,7,13,17), (2,2,2,2,2,2,2,3,3,5,7,29,31)\}$.