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\[\gcd \left( a+b, \dfrac{a^p+b^p}{a+b}\right) = 1 \text{ or } p\]

If \(p\) is an odd prime and \(a,b\) are co-prime positive integers, prove the equation above.

Note by Akshat Sharda
1 year, 9 months ago

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@Utsav Bhardwaj

Lakshya Sinha - 1 year, 9 months ago

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@Otto Bretscher @Ivan Koswara

Akshat Sharda - 1 year, 9 months ago

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Nice one Akshat !!! @Akshat Sharda

Chinmay Sangawadekar - 1 year, 9 months ago

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Just an outline since I'm at work: You can write \(\frac{a^p+b^p}{a+b}=pa^{p-1}+(a+b)q(a,b)\) for some polynomial \(q\) (simple algebra exercise). Thus a common divisor of \(\frac{a^p+b^p}{a+b}\) and \(a+b\) is also a divisor of \(pa^{p-1}\), which proves our point.

Otto Bretscher - 1 year, 9 months ago

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Exactly !!!DId the same !

Chinmay Sangawadekar - 1 year, 9 months ago

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