How to demonstrate that 59|3^(n+3)+2^(n+5)*31^n ?

3^(n + 3) = 3^n * 3^3

2^(n + 5) = 2^n * 2^5.

So our equation is 3^n * 3^3 + 2^n * 2^5 * 31^n.

27(3^n) + 32(2^n) * 31^n

Note that 59 = 32 + 27

Also, 27(3^n) will always be divisible by 27, and 32(2^n) * 31^n will also always be divisible by 32. 

Not saying that the sum will be divisible by 59, but good luck from here!