## Task 26/76

We are looking for all quadruples $\left(p_{1} ; p_{2} ; p_{3} ; p_{4}\right)$ of prime numbers that satisfy the system of equations:

$$
p_{1}^{2}+p_{2}^{2}=p_{3} \quad(1) \quad ; \quad p_{1}^{2}-p_{2}^{2}=p_{4}
$$
If both prime numbers $p_{1}$ and $p_{2}$ were odd, then $p_{4}$, as the only even prime number, could only be $p_{4}=2$, and it would be

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

This would imply: $p_{1}+p_{2}=2, p_{1}-p_{2}=1$; however, this system of equations is not solvable in integers (let alone in prime numbers). Therefore, among the prime numbers $p_{1}$ and $p_{2}$, exactly one is even and exactly one is odd.

Since $p_{4}>0$, $p_{2}$ must be $2$. Thus, (2) takes the form

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

This implies $p_{1}+2=p_{4}, p_{1}-2=1$. Therefore, $p_{1}=3, p_{4}=5$. From (1), we then get

$$
p_{1}^{2}+p_{2}^{2}=9+4=13=p_{3}
$$

The quadruple $\left(p_{1}, p_{2}, p_{3}, p_{4}\right)=(3,2,13,5)$ is thus the only one that satisfies the system of equations.