## Task $7 / 70$

We are looking for all prime numbers $p$ that satisfy the equation $c^{3}-c^{2}-11 c-1452=p$ for some positive integer $c$.
Since every prime number (except for the order) has exactly one representation as a product of two factors, $p=1-p$, it is natural to factorize the term $c^{3}-c^{2}-11 c-1452$ into a product of two factors and to investigate for which values of $c$ one of the factors takes the value 1. We therefore set

$$
c^{3}-c^{2}-11 c-1452=(c+a)\left(c^{2}+a_{1} c+a_{0}\right)=c^{3}+\left(a+a_{1}\right) c^{2}+\left(a_{0}+a a_{1}\right) c+a a_{0}
$$

By comparing coefficients, we get $a=-12, a_{1}=11, a_{0}=121$, so the desired product representation is $p=(c-12)\left(c^{2}+11 c+121\right)$. Now we need to solve the equations $c-12=1$ and $c^{2}+11+121=1$.

The first equation yields $c=13$, the second has no real solutions. Therefore, $c=13$ is the only value for which a prime number is obtained: $p=433$.