question in number theory Let $p$ is an odd prime and $n$ is an even natural number. It is clear that $2$ divides $p^n+1$. I would like to know Is the following claim true?
$4$ does not divides $p^n+1$.
 A: Let $k$ be odd and $n=2m$, and compute modulo $4$. If $k=1$ then $k^n=1$ and so $k^n+1=2$. If $k=3$, then $k^n=(k^2)^m=(3^2)^m=9^m=1^m=1$, and so $k^n+1=2$. In any case, $4$ does not divide the integer $k^n+1$ whenever $k$ is odd and $n$ is even. Primality plays no role here. 
A: $p \equiv -1$ or $1(\mod 4)$ (Why?)
$p^{2m }\equiv 1 (\mod 4) $ Since $n=2m$ 
Try proving $4|p^n+1$ if $n \in$ odd. :)
A: If $p=3, 3^n+1=(4-1)^n+1\equiv1+(-1)^n\pmod 4\implies 4|(3^n+1)$ if $n$ is odd
Other primes can be written as $(6a\pm1)$ where $a$ is any integer
Now, $(6a+1)^n+1\equiv 1+6nk+1\pmod 4\equiv2(nk+1)\pmod 4$ 
$\implies 4|\{(6a+1)^n+1\}$ if $n\cdot k$ is odd
Again, $(6a-1)^n+1\equiv(-1)^n-6nk+1\pmod 4$ 
If $n$ is odd $=2m+1$(say),$(6a-1)^n+1\equiv-1+6(2m+1)k+1\equiv2k\pmod 4$ will be divisible by $4$ if $k$ is even
If $n$ is even $=2m$(say),$(6a-1)^n+1\equiv1-6(2m)k+1\equiv2\pmod 4$ will not be divisible by $4$

Alternatively, $(2a+1)^2=4a^2+4a+1\equiv1\pmod 4$
$\implies (2a+1)^{2n}\equiv1\pmod4$
$\implies (2a+1)^{2n}+1\equiv2\pmod4$ will not be divisible by $4$
and  $(2a+1)^{2n+1}\equiv(2a+1)\pmod4$
$\implies (2a+1)^{2n+1}+1 \equiv2(a+1)\pmod4$ will be divisible by $4$ if $a$ is odd
