Prove the following (algebra of polynomials)

Just an idea for a start.

From given equation we can easily get this equation:$${P_{i+2}+P_{i}\over P_{i+1}} = {P_{i+1}+P_{i-1}\over P_{i}}$$ Since this is valid for all $i$ we have $$ {P_{i+1}+P_{i-1}\over P_{i}}= {P_{3}+P_{1}\over P_{2}} = n+1$$ and thus we have a linear equation: $$ P_{i+1} = (n+1)P_{i}-P_{i-1}$$

More generally, we have

$\qquad \gcd(P_a,P_b) = P_{\gcd(a,b)}$

a property shared by Fibonacci numbers.

Indeed, from $P_{a+1} = (n+1)P_{a}-P_{a-1}= P_2 P_{a}-P_1 P_{a-1}$, it follows by induction that $$ P_{a+b} = P_{b+1} P_{a}-P_b P_{a-1} \qquad (*) $$ and so $$ \gcd(P_{a+b}, P_b) = \gcd(P_a,P_b)= \gcd(P_{a-b},P_b) = \cdots = \gcd(P_{a-bq},P_b)=\gcd(P_b,P_r) $$ when $a=bq+r$, which reproduces the Euclidean algorithm.

The identity (*) is also reminiscent of $$ F_{a+b}=F_{b+1}F_{a}+F_{b}F_{a-1} $$ for the Fibonacci numbers.