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)

Printable View

- Nov 1st 2009, 07:14 PMMichaelGCongruence Theory Proof
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) - Nov 2nd 2009, 03:48 AMPaulRS
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 (Wink)