A Prime Diversion

We have that $n^{4} - 52n^{2} + 595 = (n^{2} - 17)(n^{2} - 35)$ . For this to be prime we must have exactly one of $n^{2} - 17$ or $n^{2} - 35$ equal to $\pm 1$ and the other term having an absolute value that is prime with a sign that allows for the product to be positive. The possible options are as follows:

• $n^{2} - 17 = -1$ when $n = \pm 4$ , for which $n^{2} - 35 = -19$ , rendering the product $19$ , a prime.

• $n^{2} - 35 = 1$ when $n = \pm 6$ , for which $n^{2} - 17 = 19$ , rendering the product $19$ , a prime.

As all $4$ suitable values for $n$ yield the same prime, we find that $P + D = 4 + 1 = \boxed{5}$ .

If $n^2=a$ and $p=a^2-52a+595$ and hence we have (by quadratic formula) $a = \dfrac{52\pm \sqrt{{52^2-2380+4p}}}{2}=26\pm \sqrt{p+81}$ which shows $p+81=k^2$ implies $\begin{cases}p = k+9 \\ 1=k-9\end{cases} \ \ \text{or} \ \ \begin{cases}p = k-9 \\ \ 1=k+9\end{cases}$ as $k>0$ thus we have $k=10$ and $p=19$ (from 1st case) which further gives $n= \sqrt{26\pm 10}= \pm 4 \ , \ \pm 6$ giving us only a single prime so $P+D=5$ .

A nice problem

n^4 - 52n^2 + 595 = (n^2 - 35)*(n^2 - 17). If the product is to be a prime, one of the factors must be 1, or -1 if the other factor is negative. We see this happens for n = 6, n = -6, n = 4, n = -4, and in all cases, the prime is 19. So P = 4 and D = 1, P + D = 5.