Find sum of two primes if their difference is equal to $3{ n }^{ 2 }-5n-1$

The difference between two prime numbers is equal to $3{ n }^{ 2 }-5n-1$. By using $n$, find the sum of them, where $n \in \mathbb{N}$.

I didn't have any idea about how can I start to solve this question .

I hope you can give me hints to help me solve this.

• Hint: $3n^2-5n-1$ is odd. – Wojowu Jan 29 '16 at 15:49

We first notice that for all integers $n,$ $3n^{2} - 5n - 1$ is odd. (You can see why this is true by noting the parity of each part.) But every prime number but $2$ is odd, and the difference between two odd numbers is even. So we know that one of the primes is $2.$ And since $2$ is the smallest prime, the other number must be $3n^{2} - 5n + 1.$ The sum of the two primes is $\boxed{3n^{2} - 5n + 3}.$
• Minor point, the question is "when is $3n^2 -5n + 1$ is prime"? In fact, some numerical results suggest surprisingly often, at least for small n. – jim Jan 29 '16 at 18:53