Prove that if $n$ is a positive odd integer then $1947\mid (46^n+296\cdot 13^n)$ This is an exercise from The Kürschák Mathematics competition from the year 1947:

Prove that if $n$ is a positive odd integer then $1947\mid (46^n+296\cdot 13^n)$.

I have the solutions in the back of the book but I would like to tackle the problem myself. I don't really know how to start, any HINTS are appreciated.
Thank you!
 A: Hint. Note that $46^2-13^2=(46+13)(46-13)=59\cdot 33=1947$ and
$$46^{2n+1}+296\cdot 13^{2n+1}=46^2\cdot 46^{2n-1}+296\cdot 13^2\cdot 13^{2n-1}
\\=13^2(46^{2n-1}+296\cdot 13^{2n-1})+(46^2-13^2)\cdot 46^n$$
A: Note that 
$$
\begin{cases}
46\equiv 1(\mod 3)\\
46\equiv 2(\mod 11)\\
46\equiv -13(\mod 59)
\end{cases}
\ \text{and}\
\begin{cases}
296\equiv -1(\mod 3)\\
296\equiv -1(\mod 11)\\
296\equiv 1(\mod 59)
\end{cases}
\ \text{and}\
\begin{cases}
13\equiv 1(\mod 3)\\
13\equiv 2(\mod 11)\\
13\equiv 13(\mod 59)
\end{cases}
$$
Hence
$$
46^n+296\cdot 13^n\equiv 1+(-1)\cdot 1=0(\mod{3})
$$
$$
46^n+296\cdot 13^n\equiv 2^n+(-1)\cdot 2^n=0(\mod{11})
$$
and 
$$
46^n+296\cdot 13^n\equiv (-13)^n+1\cdot 13^n=0(\mod{59})
$$
Thus
$$
\begin{cases}
46^n+296\cdot 13^n\equiv0(\mod{3})\\
46^n+296\cdot 13^n\equiv0(\mod{11})\\
46^n+296\cdot 13^n\equiv0(\mod{59})\\
\end{cases}
$$
Since $1947=3\cdot 11\cdot 59$, then 
$$
46^n+296\cdot 13^n\equiv0(\mod{1947})
$$
A: Because for $n=1$ it's true and for all odd $n\geq3$ we obtain:
$$46^n+296\cdot13^n=46\cdot46^{n-1}-46\cdot13^{n-1}+(46+296\cdot13)13^{n-1}=$$
$$=46\cdot(46^2-13^2)\left(46^{\frac{n-1}{2}-1}+...+13^{\frac{n-1}{2}-1}\right)+2\cdot1947\cdot13^{n-1}=$$
$$=1947\left(46\left(46^{\frac{n-1}{2}-1}+...+13^{\frac{n-1}{2}-1}\right)+2\cdot\cdot13^{n-1}\right).$$
A: Let $x_n = 46^n+296\cdot 13^n$.
Let $a= 46^n$ and $b=296\cdot 13^n$.
Then $x_{n} = a + b$ and $x_{n+2} = 46^2 a + 13^2b$.
Now $46^2 \equiv 169 = 13^2 \bmod 1947$ and so $x_{n+2} \equiv 169(a+b) = 169x_n \equiv 0 \bmod 1947$ by induction.
