## Task 28/79

All prime number pairs $(p ; P)$ are to be determined that satisfy the equation $P=14 p^{2}+1$.

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For $p=2$, $14 p^{2}+1=57$, which is not a prime number.

For $p=3$, $14 p^{2}+1=127=P$. Therefore, $(p ; P)=(3 ; 127)$ is one of the desired pairs.

If $p>3$, then $p=3 k \pm 1$ with $k \in N ; k>1$, thus

$$
14 p^{2}+1=14(3 k \pm 1)^{2}+1=14\left(9 k^{2} \pm 6 k+1\right)+1=126 k^{2} \pm 84 k+15=3\left(42 k^{2} \pm 28 k+5\right)
$$

This implies that $14 p^{2}+1$ is not a prime number. Therefore, the found pair $(p ; P)=(3 ; 127)$ is the only one.