How to expand $a^n + b^n$ for all positive exponents?

Question

This question is similar to: How to simplify $a^n - b^n$?
where the accepted answer is given as: $a^n-b^n=(a-b)\Big(\sum_{i=0}^{n-1}a^{n-1-i}b^i\Big)$

Also, when n is odd, then: $a^n + b^n = (a+b)\Big(\sum_{i=0}^{n-1}(-1)^ia^{n-1-i}b^i\Big)$ but the solution is limited to odd exponents whereas the solution above for $a^n -b^n$ accounts for all positive exponents.

Is there a solution for $a^n + b^n$ similar to $a^n -b^n$ that accounts for all positive exponents ?

Not in the reals, no. $a^2+b^2$ is irreducible there, for example. — Brian Tung, Nov 30 '21 at 21:09
ok, so the solution is irreducible for even powers. thank you — , Nov 30 '21 at 21:17
@markvs I think $a^{4k+2}+b^{4k+2}$ is divisible by $a^2+b^2$. It can be seen in the complex factorization. — Gribouillis, Nov 30 '21 at 21:26
@Gribouillis: Or, we can write $u = a^2$ and $v = b^2$, and then we're looking at $u^m+v^m$ for odd $m$ (since the OP already knows about that). — Brian Tung, Nov 30 '21 at 21:45

score 1 · Answer 1 · answered Nov 30 '21 at 21:11

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I suggest \begin{equation} a^n + b^n = \prod_{k=0}^{n-1}\left(a - b\, e^{\frac{i\pi}{n}(2k+1)}\right) \end{equation}

answered Nov 30 '21 at 21:11

Gribouillis

14,188

Is it $e^{\frac{i\pi}{n}(2k+1)}$. Did you mean k for i?. the solution is a bit abstract but I will try to work on it. – Nov 30 '21 at 21:16
I mean $i$, the usual imaginary number. – Gribouillis Nov 30 '21 at 21:17
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This comment is is for those studying #algebra-precalculus and are confused of what $i$ is. Please watch the following https://www.youtube.com/watch?v=cUzklzVXJwo . It explains how imaginary numbers were invented. Also, the solution given falls under a branch of mathematics called complex analysis so trying to use it before understanding #algebra-precalculus like I tried will not be very enjoyable. – Nov 30 '21 at 23:03
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I wouldn't say this falls under complex analysis, really you only need understand the Euler's formula and power rules (you don't need to know why Euler's formula works, and de Moivre's formula works equally well, just scrub off the $e$s) to be able to understand why the above equality holds. That being said, OP might not be expecting an answer involving complex numbers, and (as you said) a lot of people studying precalculus algebra won't know about Euler's formula. – Kyan Cheung Dec 01 '21 at 01:34

Arthur · Answer 2 · 2021-11-30T22:47:33.983

If $n=st$ with $s>1$ odd, then we have $$ a^n+b^n=(a^t)^s+(b^t)^s $$ and you can use your "$n$ is odd" factorisation with $s$ instead.

If $n$ is a power of $2$, there is no general method that I know of. Which is evidenced, for instance, by the fact that the Fermat numbers $2^{2^k}+1^{2^k}$ are sometimes prime, and it is unknown how many times that happens. We know it happens at least 5 times (for $k=0,1,2,3$ and $4$), and there are believed to be no more.

How to expand $a^n + b^n$ for all positive exponents?

2 Answers2