Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I came across a problem in Niven's number theory text (problem 51 on page 20) that asks the following:

Show that if $(a, b) = 1$ and $p$ is an odd prime, then $$\left(a + b, \frac{a^p + b^p}{a + b}\right) = 1 \text{ or } p.$$

I am not asking for a solution to this problem; instead, I'm trying to understand why $a^p + b^p$ would always be divisible by $a + b$ given the above conditions. Does anyone have any insights as to why this would be true? Where (if at all) do we use the conditions that $(a, b) = 1$ and $p$ is an odd prime?

share|cite|improve this question
Hint: $p$ is odd - try with $p$=3 (or 5) and divide through to see what happens. – Mark Bennet Oct 11 '11 at 19:19
If the gcd of a and b is not 1 or p, then the statement is obviously wrong. – Phira Oct 11 '11 at 19:46

$x^p+1$ has a zero at $x=-1$, so a factorization with factor $(x+1)$ exists (and can be given explicitly).

Now replace $x$ on both sides by $a/b$ and multiply everything with $b^p$.

share|cite|improve this answer

Explicitly, if $p$ is any odd positive integer, $$(a + b) \sum_{k=0}^{p-1} (-1)^k a^{p-1-k} b^k = a^p + b^p$$ You don't need $(a,b) = 1$, in fact $a$ and $b$ don't need to be integers (it works in any commutative ring), and you don't need $p$ to be prime.

share|cite|improve this answer

HINT $\rm\ \big(x-a,\frac{f(x)-f(a)}{x-a}\!\big) = (x-a,\:f\:\:'(a))\:$ by $\rm\: \frac{f(x)-f(a)}{x-a} \: \equiv\ f\:\:'(a)\ \ \: (mod\ \:x-a)\ $ for $\rm\ f(x)\in \mathbb Z[x]$

For further details see my post here, which elaborates on how this result is a number-theoretical analog of a well-known result about functions (polynomials), viz. about multiplicity of roots.

share|cite|improve this answer
This appears to answer what the OP stated in bold that he was not asking about. – Henning Makholm Oct 11 '11 at 21:28
@Henning In fact it's a hint for both parts of the problem and, more importantly, a link to further explanation of the conceptual aspects. That you/someone apparently downvoted for that reason is quite disturbing. It will only serve to steer the OP away from the essence of the matter. – Bill Dubuque Oct 11 '11 at 21:59
It's not very good as a hint because the OP's question does not involve the $(x-a,\cdots)$ context at all. And how did $+$ in the question become $-$ in the hint? – Henning Makholm Oct 11 '11 at 22:08
@Henning Perhaps you should think about it a bit more before making such incorrect and misleading critiques. – Bill Dubuque Oct 11 '11 at 22:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.