Can someone show me the proof that difference of like even powers of any two numbers is divisible by the sum of the bases?
Proof of even like powers?
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$\begingroup$
number-theory
divisibility
exponentiation
2 Answers
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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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0Well 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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0I need to study the $m$od operator a bit more, will get back to this, thanks for reply – 2011-11-15