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This type of question often shows up in introductory abstract math courses, computer science discrete math, or an introduction to logic course.

Show that $n^2 + n$ is even for any integer n.

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    $\begingroup$ $\textbf{Shortest argument is}$: if consider any two consecutive number one of them must be even so product will be even. $\endgroup$ Commented Oct 1, 2013 at 17:19
  • $\begingroup$ That's exactly my answer, @SiddhantTrivedi...but indeed with less words. $\endgroup$
    – DonAntonio
    Commented Oct 1, 2013 at 17:57

4 Answers 4

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$n^2+n=n(n+1)$ and either $n$ is even or $n+1$ is even. In any case, their product is even.

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  • $\begingroup$ I like how long this took compare to the suppose and expand method $\endgroup$ Commented Oct 1, 2013 at 17:11
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$$n^2+n=n(n+1)$$

So the above is the product of two consecutive integers and then one of them exactly must be even and thus so is the product.

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Before beginning, you should understand the following:

even * even = even
even + even = even
odd  + odd  = even
odd  * odd  = odd

First, suppose $n$ is odd:

n = 2k + 1
is the formula of an odd number $n$ (this means $2k+1$ always produces an odd number $n$, try it by plugging in any integer for $k$).

Let's use this fact of n = 2k + 1 with the expression we are trying to prove is always even; remember the original expression? It is: $$n^2 + n$$ is always even.

Second, plug in for $n$:

Using our odd number formula that we know n = 2k + 1
and our original expression $n^2 + n$, plug in

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

Using the right-hand side:

Expand it by "foiling": $(2k+1)^2 + (2k+1)$

$(4k^2 + 4k + 1) + (2k + 1)$

Simplified: $4k^2 + 6k + 2$

Bring back the left-hand side and put it all together:

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

Analyzing the right-hand side you can see that no matter what $k$ is, it is multiplied by 4, which makes it always even (this applies to both the $k$'s. Same goes for the $6k$ because 6 is even. And as it always has been: 2 will always be an even number.

Now you need to suppose $n$ is even, instead of odd

Do the same steps you did above but instead of n = 2k + 1 for all odd numbers, you need to know that n = 2k for all even numbers.

Take your original expression that you're trying to prove is always even: $n^2 + n$ and your newly discovered even number formula $n = 2k$ then plug in just like you did above.

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  • $\begingroup$ Ah, but if this is asked in a course on logic, don't you need to prove the four statements at the beginning of your answer? Or are they to be taken as axiomatic? $\endgroup$ Commented Oct 1, 2013 at 17:16
  • $\begingroup$ I just put that there to give a little background, some people don't think about numbers like that. $\endgroup$
    – nodebase
    Commented Oct 1, 2013 at 17:54
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Here is a different take. Use Gauss's trick:

$S = 1 + 2 + \cdots + n$

$S = n +(n-1) + \cdots + 1$

$2S = (n+1) + (n+1) + \cdots + (n+1) = n(n+1)$

Thus, $n(n+1)=2S$ is even.

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