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 ?
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For $p=2$, the number $\frac {a^p+b^p}{a+b} $ can be not an integer: for example if $a=1$ and $b=2$. Do you mean odd prime only? – Fabio Lucchini Feb 05 '19 at 06:19
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@Fabio i think it is enough to show this for atleast one p which satisfies this. – M Desmond Feb 05 '19 at 06:25