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

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

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

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

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## Comments

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TopNewest@Utsav Bhardwaj – Lakshya Sinha · 1 year, 3 months ago

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@Otto Bretscher @Ivan Koswara – Akshat Sharda · 1 year, 3 months ago

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@Akshat Sharda – Chinmay Sangawadekar · 1 year, 3 months ago

Nice one Akshat !!!Log in to reply

– Otto Bretscher · 1 year, 3 months ago

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.Log in to reply

– Chinmay Sangawadekar · 1 year, 3 months ago

Exactly !!!DId the same !Log in to reply