# $n \mid (a^{n}-b^{n}) \ \Longrightarrow$ $n \mid \frac{a^{n}-b^{n}}{a-b}$

How does one confirm that if $n \mid (a^{n}-b^{n}) \ \Longrightarrow$ $\displaystyle n \mid \frac{a^{n}-b^{n}}{a-b}$ where $a,b, n \in \mathbb{N}$.

What i idea of is to take into consideration $$(a-b)^{n} \equiv a^{n} + (-1)^{n}b^{n} \ (\text{mod} \ n)$$ and also if we intend that $n$ is weird after that we have, $$(a-b)^{n} \equiv a^{n} -b^{n} \ (\text{mod} \ n)$$ and also given that $n \mid (a^{n} - b^{n})$ we have $$(a-b)^{n} \equiv 0 \ (\text{mod} \ n)$$

I assume i am away from the verdict of the trouble, yet this is what i can work with pertaining to the trouble.

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2019-05-18 22:54:13
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Use the reality that $x \equiv y \pmod{p^{\ell}}$ (with $\ell > 0$) indicates $x^p \equiv y^p \pmod{p^{\ell+1}}$ to deal with the instance where $n$ is a prime power. Yet note that to confirm the declaration for $n$, it is adequate to confirm the declaration independently for each and every prime power separating $n$. [I am certainly omitting information, yet probably the OP could appreciate attempting to complete the illustration. ]

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2019-05-21 10:04:23
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At the heart this is actually unimportant, an effect of $\rm\ f(x) = x^n\, \Rightarrow\ f{\:'}(x)\ =\ n\, x^{n-1}\ \equiv\ 0\:\ (mod\ n).\,$ First I offer the straightforward evidence, after that I clarify the point of view in regards to by-products.

$\rm\ \ n\:|\:(a-b)\frac{a^n-b^n}{a-b}\ \Rightarrow\,\ n = m\:k,\ \ m\:|\:a-b,\ \ k\:\big|\frac{a^n-b^n}{a-b}\$ by one-of-a-kind factorization. Hence it is adequate

to show $\rm\ \ m\:|\:n,\:a-b\ \Rightarrow\ m\ \,\Big|\frac{a^n-b^n}{a-b}\: =\ a^{n-1}+a^{n-2}\:b+\:\cdots\:+a\:b^{n-2}+b^{n-1}$

But, $\rm\,\ mod\ \:m:\:\, \ a\equiv b\ \ \ \Rightarrow\, \ \ \frac{a^n-b^n}{a-b}\ \equiv\ a^{n-1}+\cdots+a^{n-1} \equiv\, n\, a^{n-1}\equiv\, 0\$ using $\rm\ m\:|\:n\quad\$ QED

The previous line is the grandfather clause $\rm\ f = x^n,\ x = b\$ of this polynomial Taylor collection approximant:

$\rm\displaystyle\quad\quad\quad\quad\quad \frac{f(x)-f(a)}{x-a} \: \equiv\ f\:'(a)\ \ \ (mod\ \:x-a)\quad$ for $\rm\ f(x)\in \mathbb Z[x]$

$\qquad$ i.e. $\rm\quad\ f(x)\ =\ f(a) +\: f\:'(a)\ (x-a) \:+\: (x-a)^2\: g(x)\quad$ for some $\rm\ g(x) \in \mathbb Z[x]$

Therefore this outcome concerning numbers is simply a grandfather clause of the adhering to well - well-known outcome concerning features (below polynomials):  an origin $\rm\ x = a\$ of $\rm\ f(x)\$ has multiplicity $\rm > 1\ \iff\ f\:'(a) \:=\: 0.\:$ In reality, like above, several outcomes concerning numbers are in fact field of expertises of outcomes concerning features. In addition, due to the fact that features have richer framework than numbers - as an example having by-products readily available - we can manipulate this framework in the function world prior to specializing to numbers. An effective instance of this is Mason is ABC theory - which has an unimportant high - college degree evidence for polynomials, yet is an unverified opinion for numbers. It generates, therefore, an unimportant evidence of FLT for polynomials. The ethical is: to confirm an outcome concerning numbers, attempt to analyze it as grandfather clause of an outcome concerning features.

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2019-05-20 22:51:45
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