Prove if $k$ is an integer, then $k^2 - 3k$ is an even integer

Question

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?

Answer 1

You have made a mistake in assigning the same variable $k$ for the odd number on the RHS of $k^2-3k=2k+1$; it should be some other letter, e.g. $k^2-3k=2l+1$.

A simpler approach would be factoring $k^2-3k=k(k-3)$. $k$, being an integer, is even or odd, and this means $k$ or $k-3$ is even respectively, so the product must always be even.

Answer 2

$k^2$ and $3k$ are either both even or both odd, hence their difference is even.

Answer 3

If $k$ is even, $k^2$ is even and $3k$ is even, so $k^2-3k$ is even.

If$k$ is odd so are $k^2$ and $3k$. $k^2-3k$ is the difference of two odd numbers, so it is even.