# Math Help - Help with proof

1. ## Help with proof

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

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

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

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

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

$4k+1=2$

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

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

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