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: Suppose for $\;n\;$ and show for $\;n+1\;$ :
$$3^{3(n+1)+3}-26(n+1)-27=27\cdot3^{3n+3}-26n-26-27=$$
$$=27\left(\color{red}{3^{3n+3}-26n-27}\right)+26\cdot26n+\overbrace{26\cdot27-26}^{=26^2}=$$
$$=27\left(\color{red}{3^{3n+3}-26n-27}\right)+26^2(n+1)$$
and the red part is divided by $\;169=13^2\;$ by the inductive assumption, and the second summand is clearly divided by $\;13^2\;$ so we have finished.
A: $A(n) = 3^{3n+3}-26n-27$   
$A(n+1) = 3^{3n+6}-26(n+1)-27$    
$A(n+1) - A(n) = 3^{3n+3}(3^3-1) - 26(n+1)-27 + 26n + 27 = 3^{3n+3} \cdot 26 - 26 = (3^{3n+3} - 1) \cdot 26 = (27^{n+1} - 1) \cdot 26$      
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 $13^2 / A(0)$ 
A: Use congruences.
$3$ has order $3$ modulo $13$, hence
$$3^{3n+3}-26n-27\equiv 1^{n+1}-0-1=0.$$
This proves it is divisible by $13$. More precisely, we have
\begin{align*}
3^{3n+3}-26n-27&=27(27^n-1)-26n=27\cdot26(27^{n-1}+27^{n-2}+\dots+27+1)-26n\\
&=2\cdot 13\bigl(27(27^{n-1}+27^{n-2}+\dots+27+1)-n\bigr).
\end{align*}
Modulo $13$, the second factor is congruent to
$$1\bigl(1(1+1+\dots+1)-n\bigr),$$
which is congruent to $0$, since there are $n$ $1$s in the sum. Hence the second factor is also divisible by 13.
A: When in doubt... just do it.
$$3^{3n+3}=(3^3)^{n+1}=(1+13\cdot2)^{n+1}\equiv1+\binom{n+1}113\cdot2\pmod{169}\equiv ?$$
