I only know the expansion of $a^p+b^p=(a+b)(a^p-1 -a^p-2×b^1......b^p-1)$ but I don't know how to proceed furthur.thankhs in advance.
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1https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – cqfd Jun 03 '19 at 08:33
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Have you tried applying the Euclidean algorithm? – Arthur Jun 03 '19 at 08:48
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https://math.stackexchange.com/questions/247146/show-that-gcd-left-fracan-bna-b-a-b-right-gcdn-dn-1-a-b – lab bhattacharjee Jun 03 '19 at 09:05
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https://math.stackexchange.com/questions/340955/gcd-leftab-fracapbpab-right-1-or-p – lab bhattacharjee Jun 03 '19 at 09:05
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1Possible duplicate of $\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1$, or $p$ – Bill Dubuque Jun 03 '19 at 14:57
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The question is not always true; if $p \mid a+b$ the $\gcd$ may not be $1$, e.g. when $(a,b,p) = (1,2,3)$ we get $\frac{a^p+b^p}{a+b} = 3 = a+b$. So we further assume that $\gcd(a+b, p) = 1$, and show that $\displaystyle \gcd \left( a+b, \frac{a^p+b^p}{a+b} \right) = 1$. Suppose that some prime $q$ divides both $a+b$ and $\frac{a^p+b^p}{a+b} = a^{p-1} - a^{p-2}b + a^{p-3}b^2 \ldots + b^{p-1}$. Since $q \mid a+b$, we have $a \equiv -b \pmod q$. So we therefore obtain $\frac{a^p+b^p}{a+b} \equiv (-b)^{p-1} - (-b)^{p-2}b + \ldots + b^{p-1} \equiv pb^{p-1} \pmod q$. But since $q \mid a+b$, and $a$ and $b$ are coprime, we have $q \not \mid b \implies q \not \mid pb^{p-1}$, and so $\frac{a^p+b^p}{a+b} \equiv pb^{p-1} \not \equiv 0 \pmod q$, yielding a contradiction.
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I think that you have done a wrong expansion of a^p+b^p÷a+b it should be a^p-1 - a^p-2×b +a^p-3×b^2 - .......+b^p-1 – Rajendra Sharma Jun 03 '19 at 09:04
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