Prove that for positive integer $ n$ we have $169| 3^{3n+3}-26n-27$. 
Prove that for positive integer $n$ we have $169| 3^{3n+3}-26n-27$.

It is easy to show that if we take the expression modulo $13$ we have $3^{n+3}-26n-27 \equiv  1^{n+1}-1 \equiv 1-1 \mod 13 = 0 \mod 13$. How do I prove it is divisible by $13^2$?

$A(n) = 3^{3n+3}-26n-27$   
$A(n+1) = 3^{3n+6}-26(n+1)-27$    
$A(n+1) - A(n) = 3^{3n+3}(3^3-1) - 26(n+1)-27 + 26n + 27 = 3^{3n+3} \cdot 26 - 26 = (3^{3n+3} - 1) \cdot 26 = (27^{n+1} - 1) \cdot 26$      
From here, it's quite obvious now, since 13 divides both multipliers.   
$13 /  (27^{n+1} - 1) $ and 
$13 / 26 $ 
Hence $13^2 / (A(n+1) - A(n))$  
Finally all you have to check is that $13^2 / A(0)$ 
