Prove the following (algebra of polynomials)

Question

Let $P_1=1$ and let $P_2=n+1$ define $$P_{i+1}=\frac{P_i^2-1}{P_{i-1}}$$ Prove that if $a \mid b$ then $ P_a \mid P_b $

I am working on this problem for a while but I could use some help here are the first 5 polynomials generated by this definition $$ P_3=n^2+2n$$ $$ P_4=n^3+3n^2+n-1$$ $$ P_5=n^4+4n^3+3n^2-2n-1$$ $$ P_6=n^5+5n^4+6n^3-2n^2-4n$$ $$ P_7=n^6+6n^5+10n^4-9n^2-2n+1$$ Note that for instance $P_2 \mid P_6$ $$ \frac{P_6}{P_2}=n^4+4n^3+2n^2−4n $$ also $P_3 \mid P_6$ $$ \frac{P_6}{P_3}=n^3+3n^2−2 $$ Finally $(P_2\times P_3) \mid P_6$ $$ \frac{P_6}{P_2 \times P_3}=n^2+2n−2 $$ So it seems to work but who can help me prove it (I think the induction method is the most appropriate) any ideas?

Answer 1

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: $$ P_{i+1} = (n+1)P_{i}-P_{i-1}$$

Answer 2

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.