Primes of the form $n^2+n+1$

Question

Let $f(n)=n^2+n+1$. While experimenting I found that

Given $m,n\in\mathbb N, \: n>1$.
If $f(2n)\in\mathbb P$ and $f((3m+1)n)\in\mathbb P$ then $3|n$.

It has ben tested for $0\le m <1000$ and $1<n<1000$. I've tried to let $n=3k+i$ looking for polynomials of $k$ for $i=1,2$ but it didn't work well.

My hope is that someone can prove the conjecture.

Answer 1

Hint: Show the following. If you're stuck, explain what you've tried.

1. $f(a) \in \mathbb{P} , a \in \mathbb{N} \Rightarrow a \equiv 0, 2 \pmod{3}$ or $ a = 1$.
   Note: Don't forget to check for the $a=1$ edge case in the following.
2. $f(2n) \in \mathbb{P} \Rightarrow n \equiv 0, 1 \pmod{3}$.
3. $f((3m+1)n )\in \mathbb{P}, n > 1 \Rightarrow (3m+1)n \equiv 0, 2 \pmod{3} \Rightarrow n\equiv 0, 2 \pmod{3}$.
4. Hence $3 \mid n$.

Answer 2

If $n\equiv 1(mod 3)$ then we have $a=n(3m+1)\equiv 1(mod 3)$.Therefore $f(a)\equiv 0(mod 3)$ and it is prime. Hence $f(a)=3$ but we know that $n>1$. Contradiction. If $n\equiv 2(mod 3)$ then we have $2n\equiv 1(mod 3)$.Therefore $f(2n)\equiv 0(mod 3)$ and it is prime. Hence $f(a)=3$ but we know that $2n>1$. Contradiction. Therefore we get $n\equiv 1(mod 3)$, which means $3\mid n$.