If $(a,b)=1$ and $n$ is a prime, prove that $ (a^n+b^n) /(a+b)$ and $(a+b)$ have no common factor, unless $a+b$ is a multiple of $n$

Question

How do I solve this without using congruence modulo or Fermat's theorems. This problem is from a book called challenges and thrills of pre college mathematics, since this problems is from first chapter of the book Fermat's theorem and congruence modulo isn't mentioned so I want to know if there is any other way of solving this problem. I know this problem is repeated but there wasn't a solution without Fermat's theorem or congruence modulo.

My approach was to solve for even primes and odd primes separately. I wasn't able to solve for even primes since $a^n+b^n$ is no

Answer 1

Hint: $$a^7+b^7= (a+b)^2[(a^5+b^5)-2ab(a^3+b^3)+3a^2b^2(a+b)]-7a^3b^3(a+b),$$ but $(a, b)=1$, so $(a+b)$ can have no prime factor in common with $a$ or $b$, so must be a multiple of 7. This generalises: (a+b)^2 is a factor of $a^{4r\pm1}+b^{4r\pm1} \mp(4r\pm1)(a+b)a^{2r}b^{2r}$.

Answer 2

Over here I converted my proof which used congruency to divisibility as they are interchangeable to a large extent.

$$a^n+b^n = (a+b)(a^{n-1}-ba^{n-2}+\cdots+b^{n-1}) \implies \frac{a^n+b^n}{a+b} = (a^{n-1}-ba^{n-2}+\cdots+b^{n-1})$$

Let $x$ be the common factor,

Then, $$x|(a^{n-1}-ba^{n-2}+\cdots+b^{n-1}) \rightarrow x|(a^{n-1}-ba^{n-2}+\cdots+b^{n-1} - a^{n-1}+a^{n-1})$$ $$\implies x|(a^{n-1}-a^{n-2}(a+b)+\cdots+b^{n-1}+a^{n-1})$$ As $x|a+b,$ we can remove $-a^{n-2}(a+b)$. $$x|(a^{n-1}+a^{n-1}+a^{n-3}b^2-\cdots+b^{n-1}+a^{n-1}-a^{n-1})$$ $$x|(a^{n-1}+a^{n-1}+a^{n-3}(b^2-a^2)-\cdots+b^{n-1}+a^{n-1})$$ As $x|(b+a)(b-a)$, we can remove $a^{n-3}(b^2-a^2)$.

$$x|(a^{n-1}+a^{n-1}+a^{n-1}-a^{n-4}b^3+\cdots+b^{n-1})$$

We can repeat adding $a^{n-1} $and subtracting it, as $x|a^{2k+1}+b^{2k+1}$ and $x|a^{2k}+b^{2k}$ due to $a+b$ being in their factorization.

At the end we get,$$x|n(a^n-1)$$

As $x|a^{n-1}-b^{n-1}$, if $x|a^n-1$ then it will have to divide $b^n-1$ also which can not happen.

Therefore, $x|n \implies x \in \{1, n\}$. If $x = n$, we must say $$n|a+b \tag*{$\blacksquare$}$$