Fermat's Theorem involving GP

Question

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.

Actually, that first statement is true for any $n$, not just those prime to $p$. — Thomas Andrews, Jul 02 '15 at 06:23
(Also, why shift from $p$ being prime and $n$ arbitrary to then using $n$ as the prime? That just makes the question hard to read. — Thomas Andrews, Jul 02 '15 at 06:25

score 1 · Answer 1 · answered Jul 02 '15 at 07:08

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 $n\nmid(a+b),(n,a+b)=1$ $$\dfrac{ab\cdot a^{n-2}+ab\cdot b^{n-2}}{a+b}\equiv1\pmod n$$

Fermat's Theorem involving GP

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