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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.

Bart Michels
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2 Answers2

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$\ (a,b)=1\Rightarrow (b,a^n\!-b^n) = (b,a^n) = 1$ so $\,\color{#c00}{b\ {\rm is\ coprime}}$ to $\,a^x\!-b^x,\,a^y\!-b^y\,$ so coprime to any common divisor $\,d.\,$ So $\,{\rm mod}\ d\!:\, c = a/b = a\color{#c00}{b^{-1}\rm\ exists}$, so $\,a^n\equiv b^n\!\!\iff\! c^n =(a/b)^n\!\equiv 1\,$ so

$$c^x\equiv 1\equiv c^y\iff {\rm ord}\, c\mid x,y\iff {\rm ord}\, c\mid(x,y)=:g\iff c^g\equiv 1\quad {\bf QED}$$

where the first & third $\!\iff\!$ is by the Order Theorem and the second is the gcd Universal Property.

Remark $ $ This method of reducing a proof about a homogeneous polynomial to a simpler nonhomogeneous case often proves handy.

Bill Dubuque
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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 $\gcd(a^x-b^x,a^y-b^y)=a^{\gcd(x,y)}-b^{\gcd(x,y)}$

  • Here's the proof using $,x,y,$ vs. $,pd,qd$

    $$\begin{align} a^x\equiv b^x, \Rightarrow&\quad\ \ \ a^{ox}\equiv b^{oy}\ \Rightarrow&\ \ a^{d+my}\equiv b^{d+my}\ \Rightarrow&\ \ a^d (a^y)^m\equiv b^d (b^y)^m\ \Rightarrow&\ \ a^d (a^y)^m\equiv b^d (a^y)^m,\ {\rm by},\ b^y\equiv a^y\ \Rightarrow&\ \ a^d \equiv b^d,\ {\rm by\ cancel},\ a^{ym} \end{align}$$ 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
  • Nice, it is very similar. $(a,n)=1$, because $(a,n)|(a,a^x-b^x)=1$ because $(a,b)=1$. I don't understand what you mean? I'm saying every $n$ that divides both $a^x-b^x$ and $a^y-b^y$ will also divide $a^d-b^d$, so $(a^x-b^x,a^y-b^y) | a^d-b^d$ but $a^d-b^d|(a^x-b^x,a^y-b^y)$ so $a^d-b^d=(a^x-b^x,a^y-b^y)$ – steedsnisps Apr 07 '15 at 12:16