Given a function $f:N\rightarrow N$ it is true that $f(1)=43$, $f(2)=142$ and $f(n+1)=3f(n)+f(n-1)$, $n\ge 2$. Prove that $\frac{f(n+1)}{f(n)}$ is irreducible.
I proved it as follows:
If $n=1$ it holds true.
I assume that it holds true for $n\le k$
$f(k+2)=3f(k+1)+f(k)$
$\frac{f(k+2)}{f(k+1)}=3+\frac{f(k)}{f(k+1)}$
Since from the assumption $\frac{f(k)}{f(k+1)}$ is irreducible, hence $\frac{f(k+2)}{f(k+1)}$ is irreducible, hence we have proved it using full induction.
Could you please look at my solution and confirm it is correct and provide other neat solutions for this question?