# Thread: Proof n^2 + n is even

1. ## Proof n^2 + n is even

Hi!

So Im not allowed to use induction on this one.
How do i prove that n^2 + n is even when n is natural number?

Not a clue on this one. Would you mind to help a bit?

2. Originally Posted by veturi
Hi!

So Im not allowed to use induction on this one.
How do i prove that n^2 + n is even when n is natural number?

Not a clue on this one. Would you mind to help a bit?
$n^2 + n = n(n+1)$. Therefore one of the factors of $n^2 + n$ is certain to be even. Therefore ....

3. oh my. so easy. Thanks a lot! I'm a bit ashamed tho...

4. A number can be represent in two forms:

n=2k with $k\in\mathbb{Z}$
or
n=2k+1 with $k\in\mathbb{Z}$

First case we have:

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

so, $2|(n^2+n)$ for all even n.

Second case:

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

so, $2|(n^2+n)$ for all odd n.

thus, $2|(n^2+n) \ \forall n\in\mathbb{Z}$

5. Originally Posted by veturi
Hi!

So Im not allowed to use induction on this one.
How do i prove that n^2 + n is even when n is natural number?

Not a clue on this one. Would you mind to help a bit?
Hello : There is the solution
I'have two cases:
1 case : $n \equiv 0\left[ {\bmod 2} \right]$

$\left\{ \begin{array}{l}
n \equiv 0\left[ {\bmod 2} \right] \\
n^2 \equiv 0\left[ {\bmod 2} \right] \\
\end{array} \right. \Rightarrow n^2 + n \equiv 0\left[ {\bmod 2} \right]$

2 case: $n \equiv 1\left[ {\bmod 2} \right]
$

$\left\{ \begin{array}{l}
n \equiv 1\left[ {\bmod 2} \right] \\
n^2 \equiv 1\left[ {\bmod 2} \right] \\
\end{array} \right. \Rightarrow n^2 + n \equiv 0\left[ {\bmod 2} \right]
$

6. Originally Posted by veturi
Hi!

So Im not allowed to use induction on this one.
How do i prove that n^2 + n is even when n is natural number?

Not a clue on this one. Would you mind to help a bit?

If $n$ is even there is nothing to prove, so assume $n$ is odd and $n^2+n$ is also odd.

As $n$ is odd it can be written $n=2k+1$ for some natural $k$, then:

$n^2+n=4k^2+4k+1+2k+1= ...$

CB

7. Originally Posted by CaptainBlack

If $n$ is even there is nothing to prove, so assume $n$ is odd and $n^2+n$ is also odd.

As $n$ is odd it can be written $n=2k+1$ for some natural $k$, then:

$n^2+n=4k^2+4k+1+2k+1= ...$

CB
Hello : you have any contradiction , because I’can written :
$\begin{array}{l}
n^2 + n = 4k^2 + 4k + 1 + 2k + 1 \\
n^2 + n = 2\left( {2k^2 + 2k + 1 + k} \right) \\
n^2 + n = 2\alpha \\
\alpha = 2k^2 + 2k + 1 + k \\
\end{array}
$

Conclusion : even number

8. Originally Posted by dhiab
Hello : you have any contradiction , because I’can written :
$\begin{array}{l}
n^2 + n = 4k^2 + 4k + 1 + 2k + 1 \\
n^2 + n = 2\left( {2k^2 + 2k + 1 + k} \right) \\
n^2 + n = 2\alpha \\
\alpha = 2k^2 + 2k + 1 + k \\
\end{array}
$

Conclusion : even number
It is a contradiction, I assume it odd and find it even hence the assumption that $n$ odd and $n^2+n$ is also odd fails, so if $n$ is odd $n^2+n$ is not odd.

CB.

CB

9. other proof
we know that
1+2+3+...+n= n(n+1)/2

so 2|n(n+1), n(n+1) is even