<s>[INST] Proof by induction that $ 169 \mid 3^{3n+3}-26n-27$ $ 169$ | $3^{3n+3}-26n-27$ ?
Fulfilled for $n=0$. Induction to $n+1$:
An integer $x$ exists so that
$ 169x= 3^{3n+6}-26n-27-26$
$ 169x= 27*3^{3n+3}-26n-27-26$
$ 169x= 26*3^{3n+3}+3^{3n+3}-26n-27-26$
An integer $m$ exists so that 
$ 169x= 26*3^{3n+3}+169m-26$
($ 13x= 2*(3^{3n+3}-1)+13m$)
Now I'm stuck although it looks simple.
Thanks for any input in advance.
 [/INST] The same, but using modules.
Let's prove by induction that $3^{3n+3} - 26n - 27 \equiv 0 \pmod {169}$
Working with module $169$:


*

*Base case. If $n = 1$, then $3^3 - 27 \equiv 0$.

*Induction. Let fix an integer $m \geq 0$ and supose that (induction hypothesis):
$$3^{3m+3}-26m -27 \equiv 0 \pmod{169}$$
or, with other words:
$$3^{3m+3} \equiv 27 + 26m \pmod{169}$$
We want to prove that
$$3^{3m+6}-26(m+1) -27 \equiv 0 \pmod{169}$$

Prove:
$$\begin{align}3^{3m+6}-26(m+1)-27&\equiv 3^3 \cdot 3^{3m+3} - 26m - 26 -27\\
&\equiv 27 \cdot 3^{3m+3} - 26m -26-27\quad(*)\\
&\equiv 27(27+26m) - 26m - 26 -27\\
&\equiv (702-26)m + (729-27-26)\\
&\equiv 4\cdot 169m + 4 \cdot 169\\
&\equiv 0m + 0 \\
&\equiv 0 \end{align}$$
We have applied induction hypothesis at $(*)$.
So, as we have proved that $3^{3n+3} - 26n - 27 \equiv 0 \pmod {169}$, then, $169 | 3^{3n+3} - 26n - 27$.
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