Proof: How to prove $n$ is odd if $n^2 + 3$ is even New to the whole proof thing.
Trying to figure out that, for all integers $n$, if $n^2 + 3$ is even, then $n$ is odd.
Thank you for the help.

If $n^2 + 3$ is even then $n^2 + 3 = 2k$ , where $k$ is an integer.
Thus, $n^2 = 2k - 3 = 2k - 4 + 1 = 2(k-2) + 1$, where $k$ is an integer, which means $k-2$ is an integer.
Now, this implies that $n^2$ is odd.
Assume $n$ is even.
Then $n$ can be written in the form $2m$, where $m$ is an integer.
Thus, $n^2 = (2m)^2 = 4m^2 = 2(2m^2)$.
We know $2m^2$ is an integer.
Thus $n^2$ is even.
But this gives us a contradiction because we already have established that $n^2$ is odd.
Therefore $n$ is odd.
