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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?

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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
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3 Answers 3

$x^p+1$ 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 $b^p$.

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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.

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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.

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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
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@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
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