Question:It is known by Fermat's Theorem that $n^p-n= M(p)$= a multiple of $p$,if $p$ is a prime number and $n$ is a prime to p.If $n$ is a prime number which divides neither $a,b$ nor $a+b$ ,prove that
$a^{n-2}.b-a^{n-3}.b^2+a^{n-4}.b^3-...+a.b^{n-2}=1+$(a multiple of $n$) ...(1)
My try:First of all I thought that $a,b,(a+b),n$ follow the Fermat's theorem by either of the below equations.
$a^n-a=$ a multiple of $n$ ....(2)
$b^n-b=$ a multiple of $n$ ....(3)
$(a+b)^n-(a+b)=$ a multiple of n .....(4)
Then I used the Sum of a GP formula to calculate the sum of LHS side of eq.(1) and I got
$a^{n-2}.b[\frac{1-(-\frac{b}{a})^{n-2}}{1-(-\frac{b}{a})}]=$ a multiple of $p$
and i solved further to get
$ab[\dfrac{a^{n-2}-b^{n-2}}{a+b}]=$a multiple of $p$ ....(5)
But I am not getting a way to relate equation to relate eq. (2),(3),(4) with eq. (5).Any help to to move towards solution will be appreciated.
For odd $n,$
$$a^{n-2}.b\left[\frac{1-(-\frac{b}{a})^{n-2}}{1-(-\frac{b}{a})}\right]=\dfrac{ab(a^{n-2}+b^{n-2})}{a+b}$$
As prime $n(>2)\nmid a, n|(a^{n-1}-1)$
$$ab\cdot a^{n-2}=b\cdot a^{n-1}\equiv b\pmod n$$
$$\implies ab\cdot a^{n-2}+ab\cdot b^{n-2}\equiv a+b\pmod n$$
As $n\nmid(a+b),(n,a+b)=1$ $$\dfrac{ab\cdot a^{n-2}+ab\cdot b^{n-2}}{a+b}\equiv1\pmod n$$