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.
 A: Hint $\ $ Conceptually, the induction is simply the first two terms of BT = Binomial Theorem:
${\rm mod}\,\ 26^2\!:\ \color{#c00}{(1\!+\!26)^{n+1}}\!\equiv 1\!+\!(n\!+\!1)26\equiv \color{#0a0}{26n\!+\!27}\,\Rightarrow\, 13^2\mid26^2\mid \color{#c00}{27^{n+1}}\!\color{#0a0}{-\!26n\!-\!27}$
Remark $ $ For completeness, below is the simple inductive proof of the first two terms of BT.
$\begin{align}{\rm mod}\,\ \color{#c00}{a^2}\!:\,\  (1+ a)^{\large n}\ \, \ \  \equiv&\,\ \ 1 + na\qquad\qquad\,\ \ \ \ \  {\rm i.e.}\ \ P(n)\\
\Rightarrow\ \ (1+a)^{\color{}{\large n+1}}\! \equiv &\  (1+na)(1 + a)\quad\ \ \ \  {\rm by}\ \ 1+a \ \ \rm times\ prior\\ 
\equiv &\,\ \ 1+ na+a+n\color{#c00}{a^2},\, \ \text{but }\ \color{#c00}{a^2\equiv 0}\ \ \rm so\\[1pt] 
 \equiv &\,\ \ 1\!+\! (n\!+\!1)a\qquad\ \ \ \ \ \ {\rm i.e.}\ \ P(\color{}{n\!+\!1}) 
\end{align}$
Generalization $ $ Using the above idea it is easy to generalize, e.g. from this answer
$\!\begin{align}\rm{\bf Theorem}\ \  \forall n\in\Bbb N\!:\ d\mid f(n) = a^n\! + b\:\!n + c  &\rm \iff d\mid \color{blue}{(a\!-\!1)^2},\, \color{brown}{a\!+\!b\!-\!1},\, \color{darkorange}{1\!+\!c}\\ &\rm \iff d\mid f(0),\,f(1),\,f(2)\end{align}$
A: This is what you should have done instead:
$$\begin{align}27\cdot3^{3n+3}-26n-27-26&=27(3^{3n+3}-26n-27)+27\cdot26n+27\cdot27-26n-27-26\\&=27\cdot169m+26\cdot26n-26\cdot26\end{align}$$
Here's the proof more clearly without unnecessary variables:
$$\begin{align}13^2&\mid3^{3(n+1)+3}-26(n+1)-27\\13^2&\mid27\cdot3^{3n+3}-26n-27-26\\13^2&\mid27(3^{3n+3}-26n-27)+26\cdot26n+26\cdot27-26\\13^2&\mid13^2\cdot4n+13^2\cdot4\end{align}$$
You use the induction hypothesis in the 3rd row.
A: I'm not sure if you are mixing up what is given and what is to be proved - perhaps you are just not writing your working very clearly.  For the induction step you have to assume that
$$3^{3n+3}-26n-27=169x$$
for some integer $x$, and then you have to prove that
$$3^{3n+6}-26n-27-26$$
is $169$ times some integer (not necessarily $x$, just some integer).  The best way to do this would be to note that by assumption
$$3^{3n+3}=169x+26n+27\ ,$$
and therefore the required expression can be written
$$\eqalign{3^{3n+6}-26n-27-26
  &=3^3(3^{3n+3})-26n-27-26\cr
  &=27(169x+26n+27)-26n-27-26\cr
  &=27\times169x+(27\times26-26)n+(27^2-2\times27+1)\ .\cr}$$
Remembering that $169=13^2$, can you see why this is $169$ times an integer?  For the "nicest" proof, see if you can do it without lots of calculations, and in particular without using a calculator.
A: 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, $169 | 3^{3n+3} - 26n - 27$.
