Is $a^n+b^n$ is divisible by $a+b$?

Question

Is $a^n+b^n$ is divisible by $a+b$ ?

This question arose when I found this post Why $a^n - b^n$ is divisible by $a-b$? in these pages.

I addressed the question as follows:

$$ a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2 -a^{n-4}b^3+\cdots\pm b^{n-1})\\ \begin{align} =a^n &-a^{n-1}b-a^{n-2}b^2+a^{n-3}b^3 -a^{n-4}b^4+\cdots\pm ab^{n-1})\\ &+a^{n-1}b+a^{n-2}b^2-a^{n-3}b^3 +a^{n-4}b^4-\cdots\mp ab^{n-1}\pm b^n)\\ =a^n\pm b^n \end{align} $$

where the +sign applies to odd $n$ and the -sign to even $n$. Therefore we find that

$$ a+b|a^n+b^n,\quad n \text{ odd}\\ a+b|a^n-b^n,\quad n \text{ even} $$ Is this correct?

a^n+b^n = (a-b)(...) The identity is wrong as written. Try it for $n=2$ for example. — dxiv, May 23 '17 at 19:40
@dxiv That was a typo. Thanks for you keen eye; I have corrected it. — Cye Waldman, May 23 '17 at 19:43
The identity is still wrong. Again, try it for $n=2,$: $;a^2+b^2=(a+b) ,\cdot, (,?,),$. — dxiv, May 23 '17 at 19:45
@dxiv Sorry, in your example $n$ is even, so $;a^2-b^2=(a+b) (a-b)$. — Cye Waldman, May 23 '17 at 19:55
Well, in the case $n$ even, you can say more: $$a^2-b^2|a^n-b^n$$ — Exodd, May 23 '17 at 19:57
@CyeWaldman Sorry, but you essentially wrote $,a^n+b^n=;\cdots;=a^n \pm b^n,$. I don't follow that. — dxiv, May 23 '17 at 19:58
@dxiv Quite correct, that cannot be true; that's why I had to split it into two results. One for each sign, in order to make it so. In other words, I started out with the assertion $a^2+b^2=\cdots$, but found out that it was something else when $n$ is even. — Cye Waldman, May 23 '17 at 20:21
If $a,b,n$ are positive integers and $n$ is even then $(a+b)|(a^n+b^n)$ iff $a=b=2^m$ for some integer $m\geq 0.$ — DanielWainfleet, May 23 '17 at 21:23
@DanielWainfleet Well, then why not $(a+b)|(a^n+b^n)$ iff $a=b$? — Cye Waldman, May 23 '17 at 21:59
Right. That's what I get for trying to do it all in my head. — DanielWainfleet, May 23 '17 at 22:31
@DanielWainfleet That's okay, you prompted me to see something i had not seen before, even though it's a degenerate case. So, thank you. — Cye Waldman, May 23 '17 at 22:41

score 2 · Answer 1 · answered May 23 '17 at 20:17

2

It is if $n$ is odd. This is a high school identity:

$$a^{2p+1}+b^{2p+1}=(a+b)\bigl(a^{2p}-a^{2p-1}b+a^{2p-2}b^2-\dots-ab^{2p-1}+b^{2p}\bigr)$$

answered May 23 '17 at 20:17

Bernard

175,478

Is $a^n+b^n$ is divisible by $a+b$?

1 Answers1