Proof by induction the divisiblity Proof by induction, that 
$$x_n=10^{(3n+2)} + 4(-1)^n\text{ is divisible by 52, when n}\in N  $$
for now I did it like that:
$$\text{for } n=0:$$
$$10^2+4=104$$
$$104/2=52$$

$$\text{Let's assume that:}$$
$$x_n=10^{(3n+2)} + 4(-1)^n=52k$$
$$\text {so else}$$
$$4(-1)^n=52k-10^{3n+2}$$

$$for \text{ n+1}:$$
$$\text {after transforms get something like that:}$$
$$52k=10^{3n+3}$$

But I'm sure, that the last step I did wrong. Actually I don't know when the proof is done, if you would help me I would be thankful.
 A: For $n+1$ you have:
$$10^{(3n+2)+3}+4(-1)^{n+1}=10^3\cdot10^{3n+2}+4(-1)^n\cdot(-1)=\\
=10^3\cdot[52k-4(-1)^n]-4(-1)^n=10^3\cdot52k-(-1)^n(4004)$$
but $4004=52\cdot 77$, then
$$10^{(3n+2)+3}+4(-1)^{n+1}=52[10^3k-77(-1)^n]$$
A: Assume that for $n\in\Bbb N$, $$10^{(3n+2)} + 4(-1)^n\text{ is divisible by 52.}  $$ This means that $$\frac{10^{(3n+2)} + 4(-1)^n}{52}\in\Bbb Z.$$ Now,
$$\begin{align}
\frac{10^{[3(n+1)+2]} + 4(-1)^{n+1}}{52}&=\frac{10^{(3n+2)}10^3 + 4(-1)^n(-1)}{52}\\
&=\frac{10^{(3n+2)}10^3 + 4(-1)^n(10^3-1001)}{52}\\
&=\frac{10^3\big[10^{(3n+2)}+4(-1)^n\big]+4(-1)^n(-1001)}{52}\\
&=10^3\cdot\frac{10^{(3n+2)} + 4(-1)^n}{52}+\frac{-4004}{52}(-1)^n\\
&=10^3\cdot\frac{10^{(3n+2)} + 4(-1)^n}{52}+(-77)(-1)^n\in\Bbb Z.
\end{align}$$
This implies that $$10^{[3(n+1)+2]} + 4(-1)^{n+1}\quad\text{is divisible by }52.$$
A: $\begin{align}{\bf Hint}\qquad\qquad  {\rm mod}\,\ 52\!:\qquad  \color{#0a0}{10^{\Large 3}} (\color{#0a0}{-4})\, &\equiv\, (\color{#0a0}{-4})(\color{#0a0}{-1})\\[.3em]
10^{\Large 3n+2} &\equiv\, (\color{}{{-}4})(-1)^{\Large n}\qquad\! {\rm i.e.}\ \ P(n) \\[.3em]
{\rm scale\ prior\ by\ 10^{\Large 3}} \Rightarrow\ \ \color{}{10^{\Large 3}}10^{\Large 3n+2} &\equiv\ \color{#0a0}{10^{\Large 3}}\,(\color{#0a0}{{-}4})(-1)^{\Large n} \\[.3em]
\Rightarrow\  10^{\Large 3(\color{#c00}{n+1})+2}\!  &\equiv (\color{#0a0}{-4})(\color{#0a0}{-1})(-1)^{\Large n}\\[.3em] 
&\equiv (-4)(-1)^{\Large \color{#c00}{n+1}}\ \ \ {\rm i.e.}\ \ P(\color{#c00}{n\!+\!1})\\
\end{align}$
Remark $\ $ More generally the same method shows
$$\begin{align} a^{\Large 2}&=\, b\\ a^{\Large3}b\, &=\, c\ (= ab^2)\\ \Rightarrow\ a^{\Large 3n+2} &=\, b\, c^{\large n}\end{align}$$
OP is the special case $\ a,b,c = 10,-4,-1\,$ in $\,\Bbb Z_52 = $ integers mod $52$
