For n>(or equal to) 1, use a congruence theory to prove 27 divides 2^(5n+1) + 5^(n+2) 43 divides 6^(n+2) + 7^(2n+1)
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Note that: $\displaystyle 2^5=32\equiv{5}(\bmod.27)$ and $\displaystyle 7^2=49\equiv{6}(\bmod.43)$ try to go on from there
Last edited by PaulRS; Nov 3rd 2009 at 03:08 AM.
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