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

Sorry about the duplicate

In another answer, however, the sum $\sum\limits_{k=0}^{n-1} (-1)^{k}a^{n-1-k}b^{k}$ was expressed as $\left(\sum\limits_{k=0}^{n-2} (-1)^{k}(k+1)a^{n-2-k}b^{k}\right)(a+b) + (-1)^{n-1}nb^{n-1}$
How was it done?

Martin Sleziak
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user137049
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  • http://math.stackexchange.com/questions/491806/proving-gcd-m-n-1 – lab bhattacharjee Mar 21 '14 at 16:16
  • is a sentence fragment. ;) – Cheerful Parsnip Mar 21 '14 at 17:38
  • My opinion is that what you have added into your question would be better asked as a new question. (Where you point out that you are asking about one particular step in and old answer and you include link to it, see this discussion on meta.) The reasons is that you already have two answers to your original questions. If you change your questions substantially, then those answers will not answer your question in its new form. (This was also discussed on meta. – Martin Sleziak Mar 22 '14 at 07:02

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Hint:

$$a^p = ((a+b)-b)^p = \sum_{k=0}^p (-1)^k\binom{p}{k} (a+b)^{p-k}b^k.$$

Daniel Fischer
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Hint $\ $ If $\,f\in\Bbb Z[x]\,$ then the Factor Theorem yields a Taylor approximant

$\quad\qquad\qquad\quad\, f(x) = f(a) +\: f\:'(a)\ (x-a) \:+\: (x-a)^2\: g(x)\quad {\rm for\ some}\ \ g(x) \in \mathbb Z[x]$

$\displaystyle\qquad\qquad\Rightarrow\ \ \frac{f(x)-f(a)}{x-a} \: \equiv\ f\:'(a)\quad ({\rm mod}\ \:x-a)$

$\quad \Rightarrow \left(x-a,\dfrac{f(x)-f(x)}{x-a}\right)\,=\, (x-a,\, f'(a))\ $ by the Euclidean algorithm

$\ \ \underset{\large \begin{array}{}\ \ x\,=\,b\\ a\ \to\ -a\end{array}}{\overset{\large f(x)\,=\,x^p}\Rightarrow} \left(b+a,\dfrac{b^p+a^p}{b+a\ }\right)\,=\!\! \underbrace{(b+a,\, pb^{p-1})}_{\large (b+a,b)\,=\,(a,b)\,=\,1}\!\! = (b+a,p) = 1\,\ {\rm or}\,\ p$

Bill Dubuque
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  • I managed to prove that their GCD can be 1 without Taylor approximant (i don't know calculus) How can i prove their GCD can be p as well ? – AdhirajSinghBrar Nov 17 '20 at 02:33