$\underline{\text { Zhenedarov R.G. }}$.

Find all such prime numbers $p, q, r$ and $s$ such that their sum is a prime number, and the numbers $p^{2}+q s$ and $p^{2}+q r$ are squares of natural numbers. (The numbers $p, q, r$ and $s$ are assumed to be distinct.)
The sum of four odd prime numbers is an even number and greater than 2. Therefore, one of our prime numbers is 2. Let's consider two cases.

1) $p=2$. Note that $p^{2}+4$ cannot be a square, so one of the remaining numbers is 2, and the others are odd. Therefore, one of the expressions $p^{2}+q s$ or $p^{2}+q r$ has the form $(2 k+1)^{2}+2(2 l+1)=4\left(k^{2}+k+1\right)+3$, which is impossible, since the square of an odd number gives a remainder of 1 when divided by 4.
2) $p=2$. Let $4+q s=a^{2}$, then $q s=(a-2)(a+2)$. If $a-2=1$, then $q s=5$, which is impossible.

Therefore, $q=a-2, s=a+2$, or vice versa, meaning the numbers $q$ and $s$ differ by 4. Similarly, $q$ and $r$ differ by 4. Thus, either $s=q-4, r=q+4$, or $r=q-4, s=q+4$. One of the numbers $q-4, q, q+4$ is divisible by 3, so $q-4=3$, meaning $q=7$, and $q+4=11$.

## Answer

$(2,7,3,11),(2,7,11,3)$