$n^2+n$ is even

Question

If $n$ is a positive integer, then $n^2+n$ is even.

$n=2k$ for some k in the integers

$n^2 +n$ = $4k^2 + 2k$ = $2(2k^2+k)$

as $2k^2+k$ is an integer $n^2+n$ is even.

Jst wondering if this proof is ok. AM i allowed to say $2k^2+k$ is an integer? Thanks

Answer 1

You have to consider a case where n is odd... n = 2k + 1

Answer 2

When you say $n = 2k$ for some integer $k$, you're assuming that $n$ is even. That was not part of the given assumptions.

But you have successfully shown that if $n$ is an even integer, then $n^2 + n$ is also even (yes, $2k^2 + k$ is an integer: if you start with an integer, then no amount of multiplication and addition will make it stop being an integer).

You have to check that it also holds for $n$ odd, i.e. $n = 2k+1$.

Answer 3

You have set $n=2k$ for some k in the integers, so your proof is ok for even $n$.

Now for odd $n$, you can do the same:

set $n=2k+1$ for some k

$n^2 +n$ = $4k^2 + 4k + 2k + 2$ = $2(2k^2+3k + 1)$

so this is even.