Proof by induction

The way you worded your post suggested that you didn't understand how it could be divisible by 2 whilst not having a 2 that can be taken out as a factor.

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The way you worded your post suggested that you didn't understand how it could be divisible by 2 whilst not having a 2 that can be taken out as a factor.

I was providing the example to show the OP that what he said wasn't necessarily true.

lol.Essentially you have 4 * (even number), so that's always divisible by 8 since it will always have a factor of 8 in it, to make that clear to the reader, you put it in the form 8g(k) so that the factor of 8 is apparent.

$k^2+k \ (k\in \mathbb{Z})$ is always divisible by 2 but doesn't have a 2 out that can be taken out as a factor seemingly.