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!

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.
