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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?

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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 $$p^2-1 = (p-1)(p+1) =3b$$

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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

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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.

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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.

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