If $\gcd(a,b)=1$, is it true that $$\gcd(a^x-b^x,a^y-b^y)=a^{\gcd(x,y)}-b^{\gcd(x,y)}\;?$$
I know that $a^{\gcd(x,y)}-b^{\gcd(x,y)}\mid a^x-b^x$ and $a^{\gcd(x,y)}-b^{\gcd(x,y)}|a^y-b^y$, so I thought of something like let $n$ be a divisor of $a^x-b^x$ and $a^y-b^y$, then $n$ must also be a divisor of $a^{\gcd(x,y)}-b^{\gcd(x,y)}$ but I am stuck.
$\ (a,b)\!=\!1\Rightarrow (b,a^n\!-b^n) = (b,a^n) \!=\! 1\,$ so $\,\color{#c00}{b\ {\rm is\ coprime}}$ to $\,a^x\!-b^x,\,a^y\!-b^y\,$ so coprime to any common divisor $\,d.\ $ So $\,{\rm mod}\ d\!:\ c \equiv \frac{a}b \equiv a\color{#c00}{b^{-1}\rm\ exists}$, $ $ so $\ a^k\equiv b^k\!\!\color{#c00}\iff\! c^k\equiv\left(\frac{a}b\right)^k\!\equiv 1\,$ so
$$\begin{align}a^x&\equiv b^x\\ a^y&\equiv b^y\end{align} \!\!\!\color{#c00}\iff\!\!\! \begin{array}{}c^x\equiv 1\\ c^y\equiv 1\end{array} \!\!\!\!\color{#0cf}{\overset{\cal O\!\!\!}\iff} {\rm ord}\, c\,|\,x,y\!\!\color{darkorange}{\overset{U\!\!\!}\iff} {\rm ord}\, c\,|\,\overbrace{(x,y)}^{g}\!\!\!\!\color{#0cf}{\overset{\cal O\!\!\!}\iff}\! c^g\equiv 1\!\!\color{#c00}\iff\! a^g\equiv b^g\quad\ $$
where above $\!\color{#0cf}{\overset{\cal O\!}\iff}$ is by the Order Theorem and $\!\color{darkorange}{\overset{U\!}\iff}$ is by the gcd Universal Property.
So $\,d\mid a^x-b^x,a^y-b^y\!\iff\! d\mid a^g-b^g,\,$ so they have the same set $\,S\,$ of common divisors $\,d,\,$ hence the same greatest common divisor (= $\max S)$.
Remark $ $ This method of reducing a proof about a homogeneous polynomial $\,a^n-b^m\,$ to a simpler nonhomogeneous polynomial $\,a^n-1\,$ often proves handy.
To prove: if $a^x \equiv b^x$ mod $n$ and $a^y \equiv b^y$ mod $n$, then $a^{gcd(x,y)} \equiv b^{gcd(x,y)}$ mod $n$.
($\gcd(a,b)=1$)
Let $d=\gcd(x,y)$, $x=p*d$, $y=q*d$, then $\gcd(p,q)=1$ so there exists $m,o > 0$ so that $po=1+qm$ or vice versa. Let $y>x$.
$(a^d)^{po} \equiv (b^d)^{po}$ mod $n$, so $(a^d)^{1+qm} \equiv (b^d)^{1+qm}$ mod $n$, so $a^d*(a^{dq})^m \equiv b^d*(b^{dq})^m$ mod $n$ so $a^d \equiv b^d$ mod $n$, so every $n$ that divides both $a^x-b^x$ and $a^y-b^y$ also divides $a^d-b^d$, so $\gcd(a^x-b^x,a^y-b^y)=a^{\gcd(x,y)}-b^{\gcd(x,y)}$
$$\begin{align} a^x\equiv b^x, \Rightarrow&\quad\ \ \ a^{ox}\equiv b^{oy}\ \Rightarrow&\ \ a^{d+my}\equiv b^{d+my}\ \Rightarrow&\ \ a^d (a^y)^m\equiv b^d (b^y)^m\ \Rightarrow&\ \ a^d (a^y)^m\equiv b^d (a^y)^m,\ {\rm by},\ b^y\equiv a^y\ \Rightarrow&\ \ a^d \equiv b^d,\ {\rm by\ cancel},\ a^{ym} \end{align}$$ Looks good, except you need to prove $,(a,n)=1,$ to be able to cancel $,a.,$ Also, you should probably say more about how you use this to get the gcd equality.
– Bill Dubuque Apr 07 '15 at 03:50