\section*{Problem 4 - 171034}

Give all prime numbers \(p\) for which \(3 p+4=z^{2}\) holds, where \(z\) is a natural number!
Assuming there is such a prime number \(p\), then with a natural number \(z\)

\[
3 p+4=z^{2} \quad \text { so } \quad 3 p=z^{2}-4=(z+2)(z-2)
\]

Since \(3 p\) and \(z+2\) are positive, \(z-2\) is also a positive integer. The only ways to factor \(3 p\) into positive integer factors are if the factors are 1 and \(3 p\) or 3 and \(p\).

Moreover, \(z+2\) and \(z-2\) differ by 4. Therefore, the factorization with 1 as a factor would lead to 5 as the second factor, which would not result in a factor of the form \(3 p\).

Thus, the only possibility is that one factor is 3 and the other is \(p=7\). Indeed, \(p=7\) satisfies the condition \(3 \cdot 7+4=25=5^{2}\). The only prime number that meets the given condition is 7.