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!

Because for $n=1$ it's true and for all odd $n\geq3$ we obtain:
$$46^n+296\cdot13^n=46\cdot46^{n-1}-46\cdot13^{n-1}+(46+296\cdot13)13^{n-1}=$$
$$=46\cdot(46^2-13^2)\left(46^{\frac{n-1}{2}-1}+...+13^{\frac{n-1}{2}-1}\right)+2\cdot1947\cdot13^{n-1}=$$
$$=1947\left(46\left(46^{\frac{n-1}{2}-1}+...+13^{\frac{n-1}{2}-1}\right)+2\cdot\cdot13^{n-1}\right).$$
