Prove that $gcd((a^{n}-b^{n})/(a-b), a-b) = gcd(n(a,b)^{n-1}, a-b)$ for a,b $\in$ $\mathbb{Z}^+$ [duplicate]

Solution 1:

$\!{\rm mod}\ \color{#c00}{a\!-\!b}\!:\ c\, := a^{n-1}\!\!+a^{n-2}b+\cdots+\!b^{n-1}\!\equiv\!\overbrace{\color{#0a0}{na^{n-1}}}^{\large {\rm by}\,\ \color{#c00}{b\,\equiv\, a}}\!\!\equiv\!\overbrace{\color{#0a0}{nb^{n-1}}}^{\large {\rm by}\,\ \color{#c00}{a\,\equiv\, b}}\!.\,$ By $\rm\color{#90f}{FD}\!=\!$ Freshman's Dream

$$\Rightarrow\ \underbrace{(a\!-\!b,c) = (a\!-\!b,\color{#0a0}{na^{n-1}\!,nb^{n-1}})}_{\Large\: (a-b,\,c)\ =\ \:\!(a-b,\,\ c\,\ {\rm mod}\,\ \color{#c00}{a-b})\ \ }\! \underset{\begin{align}\\[-2pt] \large{\rm\, by\ \, Euclid}\end{align}\qquad\qquad\qquad}{\:\! = (a\!-\!b,n(a^{n-1}\!,b^{n-1}))}\! \overset{\color{#90f}{\rm FD}}= (a\!-\!b,n(a,b)^{n-1})\,\ \ {\bf\small QED}\qquad\ \ $$

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Solution 2:

Putting $c=a-b,$ we get, $$(a-b, \frac{a^n-b^n}{a-b})=(c, \frac{(b+c)^n-b^n}c)=(c,\binom n 1 b^{n-1}+\binom n 2 b^{n-2}c+\cdots+c^{n-1})=(c,nb^{n-1})$$

As $(c,b)=(a-b,b)=(a,b)=d,$ let $\frac c C=\frac b B=d$ so that $(B,C)=1$

$$(c,nb^{n-1})=(Cd,nB^{n-1}d^{n-1})=d(C,nB^{n-1}d^{n-2})=d(C,nd^{n-2})$$ as $(B,C)=1$

$$(c,nb^{n-1})=d(C,nd^{n-2})=(Cd,nd^{n-1})=(c,nd^{n-1})=(a-b, nd^{n-1})$$

Solution 3:

We have $\large\ d=(a,b)\ ,\ $ thus $\large\ \exists\ A,B\ \ \ a=Ad,\ b=Bd,\ (A,B)=1$

$\large\left(\LARGE\frac{a^n-b^n}{a-b}\large,a-b\right)=(n d^{n-1},a-b)$
$\large\ d\left(d^{n-2}\cdot\LARGE\frac{A^{\ n}-B^{\ n}}{A-B}\large,A-B\right)=d(n d^{n-2},A-B)$

Let $\large\ m=A-B\ ,\ \ \ \ $ then $\large\ (m,B)=1$

$\large\ \left(d^{n-2}\cdot\LARGE\frac{(B+m)^n-B^n}{m}\large,m\right)=(nd^{n-2},m)$
$\large\ \left(d^{n-2}\cdot(nB^{n-1}+Qm),m\right)=(nd^{n-2},m)\ \ \ \ $ for some integer Q
$\large\ \left(nd^{n-2}B^{n-1},m\right)=(nd^{n-2},m)\ ,\ \ $ which is due to $\large (m,B)=1$.