Archive - Jan 5, 2008

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The sum of odd powers of two numbers is divisible by the sum of the two numbers

Posted January 5th, 2008 by Mgccl
in

Question 48 in Five Hundred Mathematical Challenges.

Prove that 1^{99}+2^{99}+3^{99}+4^{99}+5^{99} is divisible by 5

It's really easy when the following theorem is known.
Theorem:
a^n+b^n\equiv 0\ (mod\ a+b)
n is a positive odd integer.

The theorem is stated in the book but never showed the proof, so here is where I try to come up with the proof.
Proof:
Creating f(x) so I write less latex code.
f(x)=a^x+b^x
a odd positive integer is in the form of 2n+1, n is all non-negative integers.
\begin{eqnarray} f(2n+1)&=&f(2n-1)f(2) - a^2b^2 f(2n-3)\\ &\equiv& a^2b^2 f(2n-3)\ (mod\ a+b)\\ &\equiv& f(2n-3)\ (mod\ a+b)\\ \end{eqnarray}
Repeatedly use the formula until we have f(2n-2n+1), and it's clear that
a+b \equiv 0\ (mod\ a+b)\\ f(2n+1) \equiv 0\ (mod\ a+b)\\
Q.E.D.

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