I am trying to use congruence theorems, specifically Euler's Theorem for a proof.
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1http://math.stackexchange.com/questions/641443/proof-of-anbn-divisible-by-ab-when-n-is-odd – lab bhattacharjee Aug 11 '14 at 05:47
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$4^m + 5^m \equiv (-5)^m + 5^m \equiv -5^m + 5^m \equiv 0 \mod 9 $
AgentS
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2Just note that if we replace the symbols $4,5$ with $a,b$, and thus mod by $a+b$, the proof generalizes nicely. – Musa Al-hassy Aug 11 '14 at 18:39
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No need of them. If $m$ is an odd integer, then: $$(x+y)\mid (x^m+y^m),$$ since: $$x^m+y^m = (x+y)(x^{m-1}-\cdots+y^{m-1}).$$ Now just take $x=4$ and $y=5$.
Adriano
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Jack D'Aurizio
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We have $$4^{2m+1} = 4\cdot 16^m\equiv 4(-2)^m \pmod 9$$ and $$5^{2m+1} = 5 \cdot 25^{m} \equiv 5 (-2)^{m} \pmod{9}.$$
Add them toghether to find $4^{2m+1}+5^{2m+1} \equiv 9(-2)^m \equiv 0 \pmod9$.
Tulip
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