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Could we generalize this example of congruence issue for $x,n \in \mathbb{Z}_*$?

$$ 1+x+\cdots + x^{n-1}\equiv n \pmod {x-1} $$

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Yes, since $\ {\rm mod}\ x\!-\!1\!:\,\ x\equiv 1\,\Rightarrow\, x^2\equiv 1\,\ldots\,\Rightarrow\, x^n\equiv 1\ $ by the Congruence Product/Power Rule. – Bill Dubuque Aug 31 '14 at 14:36
up vote 2 down vote accepted

$$1 \equiv 1 \pmod{ x-1 }$$

Since $x \equiv (x-1)+1$:

$$x \equiv 1 \pmod {x-1}$$

and so:

$$x^2 \equiv 1 \pmod {x-1}$$


$$x^{n-1} \equiv 1 \pmod{x-1}$$

So,again,we have:

$$1+x+ \dots + x^{n-1} \equiv 1+1+ \dots +1 \equiv n \pmod {x-1}$$

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Yes, below are hints for $3$ proofs, using congruences, Binomial Theorem, and Taylor series.

$(1)\ \ {\rm mod}\ x\!−\!1\!:\ x\equiv 1\,\Rightarrow\,x^2\equiv 1\,\ldots\,\Rightarrow\,x^n\equiv 1\,$ by the Congruence Product/Power Rule.

$(2)\ \ x^n = (1 + x\!-\!1)^n = 1 + n(x\!-\!1) + (x\!-\!1)^2(\cdots)\,\Rightarrow\, \dfrac{x^n-1}{x-1}\equiv n\pmod{x\!-\!1}$

$(3)\ \ f(x) = f(1) + f'(1) (x-1) + (x\!-\!1)^2(\cdots)\,\Rightarrow\, \dfrac{f(x)-f(1)}{x\!-\!1}\equiv \color{#c00}{f'(1)}\pmod{x\!-\!1}$

For $\, f(x) = x^n\ $ we have $\,f'(x) = n x^{n-1}$ therefore $\,\color{#c00}{f'(1) = n}.\ $

This be viewed as a double root test, e.g. see this answer.

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