A interesting thing about exponent... does this apply to everything?

Okay... when I was solving math questions today on Brilliant, one thing just jumped out of my mind.

It is easy to see 3^{3}-1 is divisible by 2. It is not hard to see 4^{4}-1 is divisible by 3. Is 5^{5}-1 also divisible by 4? Is 6^{6}-1 also divisible by 5?

Moreover, for example, is 17^{15}-1, 17^{16}-1, 17^{17}-1, 17^{18}-1 all divisible by 16?

I think it is. Here is my proof:

Let a and n be positive integers.

a^{n} =(a-1)(a^{n-1})+a^{n-1} =(a-1)(a^{n-1}+(a-1)(a^{n-2})-(a^{n-2}) =(a-1)(a^{n-1}+(a-1)(a^{n-2})-(a^{n-2})+......(a-1)a+(a-1)+1 after factoring, we get (a-1)(a^{n-1}+a^{n-2}+a^{n-3}+......+1)+1 therefore, a^{n}-1 =(a-1)(a^{n-1}+a^{n-2}+a^{n-3}+......+1)+1-1 =(a-1)(something)

and it is divisible by a-1.

But now I have another question. Does similar theory also apply to negative integers? (for example, let n be a negative integer)

Help me please! Thank you very much!

(P.S. I only learned Algebra 1 so sometimes I could be confused to use or even look at some notations. Thank you for your tolerance!)

Note by Margaret Zheng
4 years ago

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To check, do you know what Modular Arithemtic is? How about the Binomial Theorem?

Calvin Lin Staff - 4 years ago

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I know binomial theorem, but I'm not sure with modular arithmetic. And thank you very much!

Margaret Zheng - 4 years ago

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