$n(n+1)$ is even - proof without induction

Question

Prove without induction that $n(n+1)$ is an even number for every $n \in \Bbb N$.

Answer 1

In general, one of $k$ consecutive integers must be divisible by $k$. And their product is therefore divisible by $k$ too.

Apply this with $k=2$ to get that for any $n$ there must be $2|n(n+1)$.

Answer 2

There is two cases. Either $n$ is positive and odd, or either it is positive and even. Consider the first case. Let $k$ represent an arbitrary integer.

$n=2k+1$ for $k \geq 0$

We have:

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

And because $(2k+1)(k+1)$ must be an integer, $n(n+1)$ is even.

Now for the second case:

$n=2k$ for $k \geq 1$

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

Using similar logic, it must be even.

Answer 3

Draw an $n \times (n+1)$ grid, with $n$ horizontal cells and $n+1$ vertical cells.

If $n$ is even, cover the grid with $2\times 1$ domino pieces.

If $n$ is odd, cover the grid with $1\times 2$ domino pieces.

Since the domino pieces have area $2$, the total area is even.

Answer 4

There's also the Gauss way:

$\hphantom2 S = 1 + 2 + \cdots + n$

$\hphantom2 S = n + (n-1) + \cdots +1$

$2S= n(n+1)$