## Task 13/90

Determine all solutions of the system of equations

$$
p_{1}^{2}+p_{2}^{2}=p_{3}\left(p_{1}+p_{2}\right) \quad(1) \quad ; \quad p_{1} p_{2} p_{3}-8=p_{4}
$$

in prime numbers $p_{i}(i=1 ; 2 ; 3 ; 4)$.
First, we focus on solving equation (1). It is

$$
p_{1}^{2}+p_{2}^{2}=\left(p_{1}+p_{2}\right)^{2}-2 p_{1} p_{2}=p_{3}\left(p_{1}+p_{2}\right)
$$

due to the integrality of the $p_{i}$ and because $p_{1}+p_{2}>0$, it follows that $\left.p_{1}+p_{2}\right) \mid 2 p_{1} p_{2}$. Of the 7 possible cases, 4 can be quickly ruled out:

1. $p_{1}+p_{2} \neq 2$, since $p_{1} ; p_{2} \geq 2$.
2. $p_{1}+p_{2} \neq p_{1}$, since $p_{2}>0$.
3. $p_{1}+p_{2} \neq p_{2}$, since $p_{1}>0$.
4. $p_{1}+p_{2} \neq 2 p_{1} p_{2}$, since from $p_{1}+p_{2}=2 p_{1} p_{2}$ it follows that $p_{2}=p_{1}\left(2 p_{2}-1\right)$, and thus no prime number $p_{2}$ would result due to $p_{1}>1, 2 p_{2}-1>1$.

Therefore, the remaining possibilities are:

5. $p_{1}+p_{2}=2 p_{1}$; it follows that $p_{1}=p_{2}$ and from (1) $p_{1}=p_{2}=p_{3}$;
6. $p_{1}+p_{2}=2 p_{2}$ (conclusion as in 5.);
7. $p_{1}+p_{2}=p_{1} p_{2}, p_{2}=p_{1}\left(p_{2}-1\right)$ with $p_{2}=p_{1}=p_{3}=2$ (special case of 5. and 6.),

Thus, in every possible case, $p_{1}=p_{2}=p_{3}=p$. This transforms equation (2) into the form $p^{3}-8=p^{4}$. Now,

$$
p^{3}-8=(p-2)\left(p^{2}+2 p+4\right)
$$

is a prime number only if one of the factors is 1. Since $\left(p^{2}+2 p+4\right)>1$, it can only be $(p-2)$: $p-2=1, p=3$.

Thus, $p^{3}-8=3^{3}-8=19=p_{4}$. Therefore, there is exactly one solution to the system: $p_{1}=p_{2}=$ $p_{3}=3, p_{4}=19$.