Proving Divisibility: Solving ##1900^{1990} - 1## with the Power Rule

[IDValour]

Homework Statement

Prove that $1900^{1990} - 1$ is divisible by $1991$

Homework Equations

$x^n - 1 = (x - 1)(x^{n-1} + x^{n-2} + ... + 1)$

The Attempt at a Solution

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Quite naturally the first step I took was to attempt the factorisation and see what that got me:

$1900^{1990} - 1 = (1900 - 1)(1900^{1989} + 1900^{1988} + ... + 1)$

And from here I somewhat fail to see where to go forward.

$1899$ is not divisible by $1991$ so do I need to work on the second part? If so I am having trouble seeing how to resolve it. It does seems possible to factorise the problem down to $(19*100)^{1990} - 1$ but then again this seems highly irrelevant.

Any help on this issue would be greatly appreciated.

 

[Orodruin]

Try x = -1900.

Edit: Sorry, I read divisible by 1901.

 

[IDValour]

To be honest with you, I'm beginning to think that there may have been an error and that the question was intended to read as divisible by 1901 instead of 1991. Either that or with the 1990 and 1900 swapped.

 

[Orodruin]

This might not be an unreasonable assumption. Had 1991 been a prime number you might have been able to apply Fermat's little theorem, but alas 1991 = 11*181 ...