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In Ivan Niven's book on "Introduction to the theory of numbers", there is a question in the first chapter that has been boggling me.

Given $p$ is an odd prime and $(a,b) = 1$ where $(a,b) = \gcd(a,b)$, show that $$ \left(a+b,\frac{a^p + b^p}{a+b} \right) = 1 \mbox{ or }p$$

To gain intuition, I started with $p=3$ case. I was able to prove the result by eventually showing that the above gcd was equal to $(a+b,3)$. My guess therefore is that we should be able to show that this gcd is equal to $(a+b, k(a+b)+p(u))$ where $(u,a+b)=1$. However I have not been successful in this endeavor. Perhaps there is an alternate strategy...

I would appreciate any hints/tips/ideas on how to proceed. If I get it, i'll post my method as an answer.

Gautam Shenoy
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    Write $b^p=-(-b)^p$ and use http://math.stackexchange.com/questions/247146/how-to-show-that-gcd-left-an-bn-over-a-b-a-b-right-gcdn-dn-1-a-b – lab bhattacharjee Sep 10 '14 at 16:42
  • @lab: Thanks for the reference. – Gautam Shenoy Sep 10 '14 at 16:44
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    Because you ask for *any* hint, perhaps this small exercise for myself is helpful/interesting/leading-to-somewhere for you: http://go.helms-net.de/math/expdioph/CyclicSubgroups_work.pdf It discusses the (prime)-factorization of the numerators expression as well as in relation to that of the denominator. – Gottfried Helms Sep 10 '14 at 17:28
  • Thanks Gottfried. It's very helpful and useful. Additionally, you might like the following: http://www.artofproblemsolving.com/Resources/Papers/LTE.pdf – Gautam Shenoy Sep 10 '14 at 17:33
  • Thanks Gautam - nearly the same as I'd done it but with a different flavour... I'm curious to see what's in it (and perhaps can brush my own text up) – Gottfried Helms Sep 10 '14 at 20:26

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I found a method here which was very useful.

Observe that $a^mb^n = (a+b)a^mb^{n-1} - a^{m+1}b^{n-1}$

Now $$ \frac{a^p + b^p}{a+b} = \sum_{k=0}^{p-1}(-1)^ka^kb^{p-1-k}$$

With reference to above, note that repeated application of said result will result in $$\frac{a^p + b^p}{a+b} = (a+b)M + pa^{p-1}$$ where M is some integer. This can be easily verified (I did it by taking the cases where $m$ is odd and even separately and making sure I end up with $+a^{p-1}$ in all cases).

Now since $(a,b)=1 \Rightarrow (a+b,a) = 1 \Rightarrow (a+b,a^{p-1}) = 1 $ along with $(a,bc)=(a,c)$ if $(a,b)=1$ together give $$ \left(a+b, \frac{a^p + b^p}{a+b}\right) = (a+b,pa^{p-1})=(a+b,p)$$

QED

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Gautam Shenoy
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