If $a,b$ natural numbers and $\gcd(a,b)=1$, determine all value of $\gcd(2a+b,b^5-a^5)$
Can anyone help me, I'm trying to use Euclidean algorithm but its just stuck. First,you can split $b^5-a^5=(b-a)(b^4+b^3a+b^2a^2+ba^3+a^4)$, now I want to search it one by one, already solve for $gcd(b-a,2a+b)$ but for $gcd(2a+b,b^4+b^3a+b^2a^2+ba^3+a^4)$ I still get stuck, can anyone help me ?
Let $d=\gcd(2a+b,b^5-a^5)$.
Let $e=\gcd(a,d)$.
From $e{\,\mid\,}d$, we get $e{\,\mid\,}2a+b$.
From $e{\,\mid\,}a$ and $e{\,\mid\,}2a+b$, we get $e{\,\mid\,}b$.
Thus $e{\,\mid\,}a$ and $e{\,\mid\,}b$, hence $e=1$ (since $\gcd(a,b)=1$).
From $d{\,\mid\,}2a+b$, we get $b\equiv -2a\;(\text{mod}\;d)$, hence \begin{align*} & d{\,\mid\,}b^5-a^5 \\[4pt] \implies\;& b^5-a^5\equiv 0\;(\text{mod}\;d) \\[4pt] \implies\;& (-2a)^5-a^5\equiv 0\;(\text{mod}\;d) \\[4pt] \implies\;& -33a^5\equiv 0\;(\text{mod}\;d) \\[4pt] \implies\;& 33\equiv 0\;(\text{mod}\;d) \qquad\bigl(\text{since $\gcd(a,d)=1$}\bigr) \\[4pt] \implies\;& d{\,\mid\,}33 \\[4pt] \end{align*} so $1,3,11,33$ are the only possible values of $d$.
Finally, the examples \begin{array} {|c|c|c|c|c|} \hline a&b&2a+b&b^5-a^5&d\\ \hline 1&-1&1&-2&1\\ \hline 1&1&3&0&3\\ \hline 1&9&11&59048=11{\,\cdot\,}5368&11\\ \hline 1&31&33&28629150=33{\,\cdot\,}867550&33\\ \hline \end{array} show that each of the values $1,3,11,33$ is, in fact, a realizable value of $d$.
