If $\gcd(a,b)=1$ then $$\gcd\left(\frac{a^{p}+b^{p}}{a+b},a+b\right)=1$$ unless $p|(a+b)$ Where $p$ is any prime. Use of modular arithmetic is restricted. Can you prove it just using basic divisiblity and definition of gcd ?
2026-03-26 17:53:55.1774547635
If $gcd(a,b)=1$ then $gcd\big(\frac{a^{p}+b^{p}}{a+b},a+b\big)=1$ unless $p|a+b$
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