Prove if $k$ is an integer, then $k^2 – 3k$ is an even integer
Im having some trouble with this proof. Im using proof by contrapositive (if $k^2 – 3k$ is odd integer then $k$ is not an integer), and so I set $k^2 – 3k = 2k + 1$ and rearranged it to get $k^2 -5k -1 = 0$, but I don't know what to do from here, how can I show that $k$ is not an integer?
Best Answer
$k^2$ and $3k$ are either both even or both odd, hence their difference is even.