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Can someone show me the proof that difference of like even powers of any two numbers is divisible by the sum of the bases?

2 Answers 2

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HINT: $(a^{2n} - b^{2n}) = (a^n - b^n)(a^n + b^n) =...$


Added:

In response to a comment, this proof does not use logarithms.

The expression $a^n + b^n$ is called "the sum of nth powers". Notice that "the sum of the bases" is just $(a + b)$, which is the first factor in the factorization of the sum of nth powers.

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    He probably needs a second hint, namely that different things happen when $n$ odd and $n$ even...2011-11-15
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    User, see hint #2, above! (@N. S. indeed, I was hoping to evoke some interaction with our first-time OP)2011-11-15
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    I can delete my hint if you prefer :)2011-11-15
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    @N. S. interacting with *anyone* is fine with me! ;)2011-11-15
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    I don't know why people voted my hint more, your hint is actually cleaner and more elementary :)2011-11-15
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    I strongly believe in the "rich get richer" phenomenon on MsE... so after I gave yours the first vote, it was all uphill! It's not a big deal to me. I get random down votes on old questions and answers, and I've authored a top 5 (6?) answer (from an old account). I just hope people learn math and enjoy the process.2011-11-15
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    Still not getting it guys =/. Two questions. Does the proof involve logarithms? Isn't the above expression in the most simplified form? I was thinking... I need to show that (a^n - b^n)(a^n+b^n) = (a+b)i, then I took the log of both sides and got nowhere2011-11-15
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    Well if you ever return to MSE, I hope you find this solution more fulfilling!2012-02-17
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Alternate Hint: $a \equiv-b \pmod{a+b}$

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    You want `\equiv` instead of `\cong`...2011-11-15
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    I need to study the mod operator a bit more, will get back to this, thanks for reply2011-11-15