11n+2 + 122n+1
for n = 1 we get
112 + 123 = 1331 + 1728 = 3058
3059/133 = 23
hence it is true for n = 1
let us assume that this expression is true for n = k
11k+2 + 122k + 1 is divisible by 133 for n = k + 1
11k + 3 + 122k + 3 = (11k x 1331) + (122k x 1728) = (11k x 1331) + (11 + 1)2k x 1728)
which is divisible by 1331 + 1728 = 3059 which is divisible by 133.
Hence our assumption is correct and 11n+2 + 122n+1 is divisible by 133.