Show that $17$ divides $6^{3n+2} - 2\cdot 29^n$ for all natural numbers $n$ 
Show that $17$ divides $6^{3n+2} - 2\cdot 29^n$ for all natural numbers $n$.

I know that if $$17 \mid 6^{3n+2} - 2\cdot 29^n$$ 
then $6^{3n+2}$ is congruent to $2\cdot 29^n$ mod $17$.
But how can I then prove that $6^{3n+2}$ is congruent to $2\cdot 29^n$ mod $17$? Could the fact that $17$ is prime be helpful in that Fermat's Little Theorem could be used? Or should I be considering indice rules?
 A: Do arithmetic modulo $\;17\;$ all the time:
$$6^{3n+2}-2\cdot29^n=6^2\left(6^2\cdot 6\right)^n
-2\cdot 12^n=2(12)^n-2(12)^n=0$$
A: Hint $\ {\rm mod}\ 17\!:\ \color{#c00}{6^2\equiv 2}\,\Rightarrow\,\color{#0a0}{6^3\equiv 12}$   
Hence $\ 6^{2+3n} \equiv \color{#c00}{6^2}(\color{#0a0}{6^3})^n\equiv \color{#c00}2(\color{#0a0}{12})^n\equiv 2(29)^n,\,$ by $\,12\equiv 29$.
Above we implicitly employed the following fundamental rules of congruence arithmetic.

Congruence Sum Rule $\rm\qquad\quad  A\equiv a,\quad B\equiv b\ \Rightarrow\ \color{#c0f}{A+B\,\equiv\, a+b}\ \ \ (mod\ m)$
Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a) + (B\!-\!b)\ =\ \color{#c0f}{A+B - (a+b)} $
Congruence Product Rule $\rm\quad\ A\equiv a,\ \ and \ \  B\equiv b\ \Rightarrow\ \color{blue}{AB\equiv ab}\ \ \ (mod\ m)$
Proof $\rm\ \ m\: |\: A\!-\!a,\ B\!-\!b\ \Rightarrow\ m\ |\ (A\!-\!a)\ B + a\ (B\!-\!b)\ =\ \color{blue}{AB - ab} $
Congruence Power Rule $\rm\qquad \color{}{A\equiv a}\ \Rightarrow\ \color{#c00}{A^n\equiv a^n}\ \  (mod\ m)$
Proof $\ $ It is true for $\rm\,n=1\,$ and $\rm\,A\equiv a,\ A^n\equiv a^n \Rightarrow\, \color{#c00}{A^{n+1}\equiv a^{n+1}},\,$ by the Product Rule, so the result follows by induction on $\,n.$
Polynomial Congruence Rule $\ $ If $\,f(x)\,$ is polynomial with integer coefficients then  $\ A\equiv a\ \Rightarrow\ f(A)\equiv f(a)\,\pmod m.$
Proof $\ $ By induction on $\, n = $ degree $f.\,$ Clear if $\, n = 0.\,$ Else $\,f(x) = f(0) + x\,g(x)\,$ for $\,g(x)\,$ a polynomial with integer coefficients of degree $< n.\,$  By induction $\,g(A)\equiv g(a)\,$ so $\, A g(A)\equiv a g(A)\,$ by the Product Rule. Hence $\,f(A) = f(0)+Ag(A)\equiv f(0)+ag(a) = f(a)\,$ by the Sum Rule. 
Beware $ $ that such rules need not hold true for other operations, e.g.
the exponential analog of above $\rm A^B\equiv a^b$ is not generally true (unless $\rm B = b,\,$ so it reduces to the Power Rule, so follows by inductively applying $\,\rm b\,$ times the Product Rule).
