<s>[INST] 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 $13^2$?
 [/INST] 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.
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