Prove by case; if $n$ is an odd integer, then there is an integer $m$ such that $n=4m+1$ or $n=4m+3$

Question

I'm not sure how the formula $n=2k+1$ (for odd) or $n=2k$ (for even) work in this case due to the variable $m$.

Please help, Thanks.

Answer 1

Hint:

Separate the cases based on whether $k$ is odd or not.

Answer 2

Hint: Write $n=4m+r$ with $r=0,1,2,3$. Which choices of $r$ give $n$ odd?