## Task 20/81

Determine all natural numbers $n$ for which $n^{3}+1=4 p$ holds; here $p$ denotes a prime number.
The given equation is equivalent to the equation $p=0.25\left(n^{3}+1\right)$. For $p$ to be an integer, it must be that $n^{3} \equiv-1(\bmod 4)$, and thus also $n \equiv-1(\bmod 4)$. We therefore set $n=4 t-1$ with $t \in \mathbb{N}$. Then we have

\[
\begin{aligned}
& p=0.25\left(n^{3}+1\right)= 0.25\left[(4 t-1)^{3}+1\right]=0.25\left(64 t^{3}-48 t^{2}+12 t-1+1\right)= \\
& 16 t^{3}-12 t^{2}+3 t=t\left(16 t^{2}-12 t+3\right)
\end{aligned}
\]

If $p$ is to be a prime number, it is necessary that one of the two factors is equal to 1.

1. Let $t=1$. Then $12-12+3=7$ is indeed a prime number $p$, and we have $n=4 t-1=3$. Thus, a solution is found.
2. Let $16 t^{2}-12 t+3=1$. This (quadratic in $t$) equation does not yield integer values for $t=p$. Consequently, the pair $n=3, p=7$ is the only solution.