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.

  • 4
    $\begingroup$ $\textbf{Shortest argument is}$: 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


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

  • $\begingroup$ I like how long this took compare to the suppose and expand method $\endgroup$ Commented Oct 1, 2013 at 17:11


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.


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.

  • $\begingroup$ 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

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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