A number theory problem by ppk k

By observation we note that $n = 1$ is a root of the given expression, so factoring out $n - 1$ we have that

$n^{2} - 8n^{2} + 20n - 13 = (n - 1)(n^{2} - 7n + 13)$ .

For this to be prime we must have either $n - 1 = \pm 1$ and $n^{2} - 7n + 13 = \pm p$ for some prime $p$ , or $n^{2} - 7n + 13 = \pm 1$ and $n - 1 = \pm p$ for some prime $p$ . So looking at the 4 cases:

• $n - 1 = 1 \Longrightarrow n = 2$ , for which $n^{2} - 7n + 13 = 3$ , which is prime, so $n = 2$ is a solution;

• $n - 1 = -1 \Longrightarrow n = 0$ , which is not positive;

• $n^{2} - 7n + 13 = 1 \Longrightarrow (n - 3)(n - 4) = 0 \Longrightarrow n = 3$ or $n = 4$ , and since $n - 1$ is prime in both of these cases we have that $n = 3$ and $n = 4$ are solutions;

• $n^{2} - 7n + 13 = -1 \Longrightarrow n^{2} - 7n + 14 = 0$ , which does not have real roots as $7^{2} - 4 \times 14 \lt 0$ .

Thus $2,3,4$ are the only suitable values for $n$ , giving us a final answer of $\boxed{3}$ .

The above cubic factors into $n^3 - 8n^2 + 20n - 13 = (n-1)(n^2 - 7n + 13).$ If it is to equal a prime number $p$ , then we require:

$n - 1 = 1; n^2 - 7n + 13 = p \Rightarrow \boxed{n = 2, p = 3}$

$n - 1 = p; n^2 - 7n + 13 = 1 \Rightarrow \boxed{n = 3, p = 2}$ and $\boxed{n = 4, p = 3}$

Hence, there are three possible positive integers $n = 2,3,4.$

You just find a factor of the above expression And check the values. Might be the sufficient hint!!