# Find all prime number $p$ such that $2^p+p^2$ is prime number

Find all prime number $$p$$ such that $$2^p+p^2$$ is prime number

First I try with $$p=2$$, that is not true because I get $$8$$ and that is not prime number. Then I try with $$p=3$$ that is true, I get $$17$$. Now I need to see all prime number that is bigger then $$3$$. Every prime number $$p>3$$ can write as $$p=6k+1$$ or $$p=6k-1$$, $$k\in \mathbb N$$.

I try to put $$p=6k+1$$ then I get $$p^{6k+1}+(6k+1)^2=2(2^{3k})^2+36k^2+12k+1$$ then I try to factorization this like that $$a=2^{3k}$$ and $$b=6k$$ then I get $$2a^2+b^2+2b+1$$ I change because I think it is easy to see how to factorization, but there I stop I do not how to continue, I do not have idea how to factorization to get something like this $$x\cdot y=n$$, where $$n$$ is prime number, and then since every prime number is $$n=1\cdot n$$ so one of them x or y must be 1 that is my idea for the task.

Can you help me?

## 4 Answers

If $$p$$ is odd, then you can do this $$2^p+p^2 = (2^p+1)+(p^2-1)$$

which is multiple of 3 and bigger than 3 if $$p\ne 3$$.

Detailed:

In that case $$2^p+1 = (2+1)(2^{p-1}-2^{p-2}+...+1) = 3\cdot a$$ and since among three consecutive integers one is always divisible by $$3$$ and $$p\ne 3$$ we have $$p^2-1 = (p-1)(p+1) =3b$$

If $$p$$ is a prime number and $$p>3$$, then $$2^p+p^2 \equiv 2+1 \equiv 0$$ $$mod$$ $$3$$, and $$2^p+p^2>3$$, hence $$2^p+p^2$$ won't be a prime number, since it is a mltiple of $$3$$. Thus according to your argument, $$p=3$$ is the only solution

If you put p = 3 ; you get 17 which is a prime. For any prime p > 3 ; 2^p = 2 ( mod 3 ) and p^2 = 1 ( mod 3 ) and hence the sum of them must divisible by 3.

You can easily show that $$2^n \equiv 2$$ (mod 6) for each odd number $$n$$. So for an odd prime number $$p>3$$:

$$2^p\equiv 2, \quad p\equiv\pm1,\quad p^2=1\quad\implies\quad 2^p+p^2\equiv3\mod6$$

So the number cannot be prime for and $$p$$>3, it's always divisible by 3.