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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?

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  • $\begingroup$ $x^3$ 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
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Your example is not proving the statement wrong, because your calculation is wrong:

For

$$ x^3+x $$

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
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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.

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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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