# Thread: Prove x^2 + x is even?

1. ## Prove x^2 + x is even?

Prove that for every integer x, $x^2 + x$ is even.

This is what i have so far:

Suppose $x^2 + x$ is ODD

$x = 2k + 1$ (where k is any integer)

now plug in (2k + 1) for $x^2 + x$ is ODD

$(2k + 1)^2 + (2k + 1)$ = ODD

$4k^2 + 4k + 1 + 2k + 1$ = ODD

$4k^2 + 6k + 2$ = ODD

This is False!, ex. when k = 1, $4k^2 + 6k + 2$ is EVEN

Therefore $x^2 + x$ is EVEN

Does this all seem correct? i am very new to proofs. any criticism is appreciated

Thank you

2. Originally Posted by mybrohshi5
Prove that for every integer x, $x^2 + x$ is even.

This is what i have so far:

Suppose $x^2 + x$ is ODD

$x = 2k + 1$ (where k is any integer)

now plug in (2k + 1) for $x^2 + x$ is ODD

$(2k + 1)^2 + (2k + 1)$ = ODD

$4k^2 + 4k + 1 + 2k + 1$ = ODD

$4k^2 + 6k + 2$ = ODD

This is False!, ex. when k = 1, $4k^2 + 6k + 2$ is EVEN

Therefore $x^2 + x$ is EVEN

Does this all seem correct? i am very new to proofs. any criticism is appreciated

Thank you
I don't think what you have follows. If I read it right you assume by way of contradiction that

$x^2+x$ is odd this implies that
$x^2+x=2k+1$ for some $k \in \mathbb{Z}$

This does not imply that $x=2k+1$.

Why not look at it this way

$x^2+x=x(x+1)$

Now just look at two cases
what x is even $x=2k$
and what if x is odd $x=2k+1$

either way notice that no matter what $x$ is either $x$ or $x+1$ is even and so is there product.

3. Here is another way.
For any two consecutive integers, one is even the other is odd.
Even times an odd is even.

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