0
$\begingroup$

The condition on $n\in\mathbb N^+$ for $$\forall p\in\mathbb P:\, p^2+2n\notin\mathbb P$$ seems to be: $$n\equiv 1\pmod 3\wedge 9+2n\notin\mathbb P$$

I would like to see a proof or a counter-example.

$\endgroup$
  • $\begingroup$ I believe you should edit your post and omit $\forall p\in\Bbb{P}$. $\endgroup$ – Galc127 Mar 5 '17 at 13:13
  • $\begingroup$ @Galc127: You are right. Have edited. $\endgroup$ – Lehs Mar 5 '17 at 13:28
2
$\begingroup$

If $\,p\in\mathbb P\,$ and $\,p\neq3$, then $$p\equiv1\ \ ({\rm mod}\ 3)\quad\text{or}\quad p\equiv2\ \ ({\rm mod}\ 3)$$ Square both sides and we have $$p^2\equiv1\ \ ({\rm mod}\ 3)$$

First, if $\ \,n\equiv1\ \ ({\rm mod}\ 3)$, $\ \ $then $$p^2+2n\equiv1+2\equiv0\ \ ({\rm mod}\ 3)$$ Thus, $\,p^2+2n\,$ is divisible by $\,3$, which means that $\,p^2+2n\notin\mathbb P$

Besides, if $\,p=3\,$ then for $\,p^2+2n\notin\mathbb P$, it requires $$\,p^2+2n=9+2n\notin\mathbb P$$

Thus, we have proven that if $$n\equiv1\ \ ({\rm mod}\ 3)\quad\text{and}\quad9+2n\notin\mathbb P$$ then $\,p\in\mathbb P\ \Rightarrow\ p^2+2n\notin\mathbb P\,$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.