2011/034) show that 3^2n+ 3^n + 1 is divisible by 13 if n is not multiple of 3

proof:

we know 3^3 = 27 = 1 mod 13

so we proceed

if n is of the form 3k + 2 then 2n of form 3m + 1 and if n is of the form 3k + 1 the 2n of the form 3m + 2

so we have
3^2n+ 3^n + 1 = 3^ (3k + 2) + 3^(3m + 1) + 1

= 27^k * 9 + 27^m * 3 + 1
= 9 + 3 +1 mod 13 = 13 mod 13 = 0 mod 13 do divisible by 13