## Task 10/89

At least two of the four prime numbers $p_{1} ; p_{2} ; p_{3} ; p_{4}$ with $p_{1}<p_{2} ; p_{3} ; p_{4}$ are to be determined, for which $P=p_{1}^{p_{2}}+p_{3}^{2 p_{4}}$ can be a prime number.
If $p_{1}$ and $p_{3}$ are both odd, then $P$ is always even and, since $P>2$, it is not a prime number. Consequently, because $p_{1}<3$. Then,

$$
P=2^{p_{2}}+p_{3}^{2 p_{4}} \equiv(-1)^{p_{2}}+( \pm 1)^{2 p_{4}} \equiv(-1)^{p_{2}}+(+1)^{p_{4}} \equiv 0 \quad(\bmod 3)
$$

since $p_{2} ; p_{4} \geq 3$, and thus are odd. Since $P>3$, $P$ is not a prime number.
2. $p_{3}=3$. Then,

$$
P=2^{p_{2}}+3^{2 p_{4}}=2^{p_{2}}+9^{p_{4}}
$$

It follows that: If the prime quadruplet $(p 1 ; p 2 ; p 3 ; p 4)$ with $p_{1}<p_{2} ; p_{3} ; p_{4}$ is to yield its prime number $P=p_{1}^{p_{2}}+p_{3}^{2 p_{4}}$, then $p_{1}=2$ and $p_{3}=3$ must hold. Indeed, with $p_{2}=5, p_{4}=3$, the prime number $P=2^{5}+3^{2 \cdot 3}=761$ is obtained.