1. ## Help with proof

If $\displaystyle n^2+1$ is prime, then $\displaystyle n^2+1$ can be expressed as $\displaystyle 4k+1$

I was thinking you probably take $\displaystyle n^2+1$ and maybe do $\displaystyle n^2+1=p$ and manipulate it from there?

2. Originally Posted by JSB1917
If $\displaystyle n^2+1$ is prime, then $\displaystyle n^2+1$ can be expressed as $\displaystyle 4k+1$

I was thinking you probably take $\displaystyle n^2+1$ and maybe do $\displaystyle n^2+1=p$ and manipulate it from there?
Unless I am missing something the statement is false.

let $\displaystyle n=1 \implies 1^2+1=2$ is prime but there isn't an integer such that

$\displaystyle 4k+1=2$

3. Originally Posted by JSB1917
If $\displaystyle n^2+1$ is prime, then $\displaystyle n^2+1$ can be expressed as $\displaystyle 4k+1$
That is not true if $\displaystyle n=1$. Is it?
2 is prime.

4. sorry, I forgot to put n =/= 1.

5. If $\displaystyle n>0~\&~n^2+1$ is a prime number, then it is odd.
So $\displaystyle n^2$ is even. WHY?
That means $\displaystyle n$ is even, or $\displaystyle n=2k$ for some $\displaystyle k$
Therefore, $\displaystyle n^2+1=4j+1$ where $\displaystyle j=k^2.$