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

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

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

This is what i have so far:

Suppose $\displaystyle x^2 + x$ is ODD

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

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

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

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

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

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

Therefore $\displaystyle x^2 + x$ 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, $\displaystyle x^2 + x$ is even.

This is what i have so far:

Suppose $\displaystyle x^2 + x$ is ODD

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

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

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

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

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

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

Therefore $\displaystyle x^2 + x$ 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

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

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

Why not look at it this way

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

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

either way notice that no matter what $\displaystyle x$ is either $\displaystyle x$ or $\displaystyle x+1$ 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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# PROVE THAT X^3 X^2 IS EVEN

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