I have been stuck on the statement for a while; is it true or false?

Prove or disprove: $x^3 + x$ is even For all integers $x$.

If we let $x = 2$, we get 9 which is an odd number...is this enough to prove it wrong?

  • $\begingroup$ $x^3$ is odd for any odd number and $x^3$ is even for each even number. $\endgroup$ – Offlaw Jan 31 at 3:21
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    $\begingroup$ $2^3+2=8+2=10$. $10$ is even. $\endgroup$ – Gnumbertester Jan 31 at 3:22

Your example is not proving the statement wrong, because your calculation is wrong:


$$ x^3+x $$

you get

$$ 2^3 +2 = 8 +2 =10 $$

In order to prove the statement you need to prove the two statements:

  • If $x$ is odd $\Rightarrow x^3$ is odd. And thus $x^3+x$ would be (odd+odd)=even

  • If $x$ is even $\Rightarrow x^3$ is even. And thus $x^3+x$ would be (even+even)=even

Can you proceed from here?

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    $\begingroup$ my bad! i calculated it wrong...thanks bud!! $\endgroup$ – player99 Jan 31 at 3:36

A possibly bit simpler way to handle this is to note that $x^3 + x = x\left(x^2 + 1\right)$. Thus, either $x$ is even or it's odd, in which case $x^2 + 1$ is even. In either case, the product is even.


The best way to formally prove this is test what happens when you substitute in all even and of numbers. The definition of an even number $x$ is

$$x = 2s$$

And an odd number can be viewed as

$$x = 2s + 1$$

If you substitute in these values for $x$, what does the output look like in both cases? If they are both even, then for every $x$, your function will be even. Otherwise, it will not.


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