Consider a sequence of polynomials $ P_{0}(x), P_{1}(x), P_{2}(x),\ldots, P_{n}(x),\ldots $, where $ P_{0}(x) = 2, P_{1}(x) = x $ and for every $ n\geq 1 $ the following equality holds: \[ P_{n+1}(x)+P_{n-1}(x) = xP_{n}(x). \] Prove that there exist three real numbers $ a, b, c $ such that for all $ n\geq 1 $, \[ (x^{2}-4)[P_{n}^{2}(x)-4] = [aP_{n+1}(x)+bP_{n}(x)+cP_{n-1}(x)]^{2}. \] \[OOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO\] With $ P_{n+1}(x)+P_{n-1}(x) = xP_{n}(x)$, we find: \[P_{2}(x)=x^{2}-2\] \[P_{3}(x)=x^{3}-3x\] \[P_{4}(x)=x^{4}-4x^{2}+2\] We try $ (x^{2}-4)[P_{n}^{2}(x)-4] = [aP_{n+1}(x)+bP_{n}(x)+cP_{n-1}(x)]^{2} $ for $ n = 1 $: \[ (x^{2}-4)[P_{1}^{2}(x)-4] = [aP_{2}(x)+bP_{1}(x)+cP_{0}(x)]^{2} \] \[ (x^{2}-4)(x^{2}-4) = [a(x^{2}-2)+bx+c2]^{2} \] \[ (x^{2}-4)^{2} = [a(x^{2}-2)+bx+c2]^{2} \] We see $ a = 1, b = 0, c = -1 $ We try again $ (x^{2}-4)[P_{n}^{2}(x)-4] = [P_{n+1}(x)-P_{n-1}(x)]^{2} $ for $ n = 2 $: \[(x^{2}-4)[P_{2}^{2}(x)-4] = [P_{3}(x)-P_{1}(x)]^{2}\] \[(x^{2}-4)[(x^{2}-2)^{2}-4] = [x^{3}-3x-x]^{2}\] \[(x^{2}-4)(x^{4}-4x^{2}+4-4) = [x^{3}-4x]^{2}\] \[(x^{2}-4)(x^{4}-4x^{2}) = [x(x^{2}-4)]^{2}\] \[(x^{2}-4)x^{2}(x^{2}-4) = [x(x^{2}-4)]^{2}\] We prove $ (x^{2}-4)[P_{n}^{2}(x)-4] = [P_{n+1}(x)-P_{n-1}(x)]^{2} $ for $ n \geq 3$ Because $P_{n+1}(x) = xP_{n}(x)-P_{n-1}(x)$, \[[P_{n+1}(x)-P_{n-1}(x)]^{2}=[xP_{n}(x)-P_{n-1}(x)-P_{n-1}(x)]^{2}\] \[[P_{n+1}(x)-P_{n-1}(x)]^{2}=[xP_{n}(x)-2 \cdot P_{n-1}(x)]^{2}\] \[[P_{n+1}(x)-P_{n-1}(x)]^{2}=x^{2}P_{n}^{2}(x)-4 x P_{n}(x) P_{n-1}(x)+4 P_{n-1}^{2}(x)\] Because $ x P_{n-1}(x) = P_{n}(x)+P_{n-2}(x)$, \[[P_{n+1}(x)-P_{n-1}(x)]^{2}=x^{2}P_{n}^{2}(x)-4 P_{n}(x)(P_{n}(x)+P_{n-2}(x) )+4 P_{n-1}^{2}(x)\] \[[P_{n+1}(x)-P_{n-1}(x)]^{2}=(x^{2}-4)P_{n}^{2}(x)+4 (P_{n-1}^{2}(x)-P_{n}(x)P_{n-2}(x))\] \[[P_{n+1}(x)-P_{n-1}(x)]^{2}=(x^{2}-4)P_{n}^{2}(x)+4 (4-x^{2})\cdots\cdots(*)\] \[[P_{n+1}(x)-P_{n-1}(x)]^{2}=(x^{2}-4)(P_{n}^{2}(x)-4)\] q.e.d. Only to prove is (*): $ P_{n-1}^{2}(x)-P_{n}(x)P_{n-2}(x)= 4-x^{2}$ for $ n \geq 3$ Prove by mathematical induction. a) $ P_{n-1}^{2}(x)-P_{n}(x)P_{n-2}(x)= 4-x^{2}$ for $ n = 3 $ \[P_{2}^{2}(x)-P_{3}(x)P_{1}(x)= (x^{2}-2)^{2}-(x^{3}-3x)x\] \[= (x^{4}-4x^{2}+4-x^{4}+3x^{2})\] \[= (x^{4}-4x^{2}+4-x^{4}+3x^{2})\] \[= 4-x^{2}\] b)Suppose: $ P_{n-1}^{2}(x)-P_{n}(x)P_{n-2}(x)= 4-x^{2}$ for $ n $ To prove: $ P_{n-1}^{2}(x)-P_{n}(x)P_{n-2}(x)= 4-x^{2}$ for $ n + 1$ \[P_{n}^{2}(x)-P_{n+1}(x)P_{n-1}(x)\] \[=P_{n}^{2}(x)-[x P_{n}(x)-P_{n-1}(x)]P_{n-1}(x)\] \[=P_{n}^{2}(x)-x P_{n}(x)P_{n-1}(x)+P_{n-1}^{2}(x)\] \[=P_{n}^{2}(x)-x P_{n}(x)P_{n-1}(x)+(4-x^{2})+P_{n}(x)P_{n-2}(x)\]see suposition \[=P_{n}(x)[P_{n}(x)-x P_{n-1}(x)+P_{n-2}(x)]+(4-x^{2})\] \[=P_{n}(x)\cdot (0)+(4-x^{2})\] \[=4-x^{2}\]

One can prove by induction on $n$ that $P_n(2\cos t) = 2\cos(nt)$ for all $t$, so if $x = 2\cos t$, then $$\begin{aligned}(x^2-4)\left(P_n(x)^2 - 4\right) &= 16(\cos^2t - 1)\left(\cos^2(nt)-1\right)\\&= \left(-4\sin t\sin(nt)\right)^2\\&= \left(2\cos(n+1)t - 2\cos(n-1)t\right)^2\\&= \left(P_{n+1}(x) - P_{n-1}(x)\right)^2. \end{aligned}$$As $t$ varies, $x = 2\cos t$ can take infinitely many values, so the polynomial equation $$ (x^2-4)\left(P_n(x)^2 - 4\right) = \left(P_{n+1}(x) - P_{n-1}(x)\right)^2 $$has infinitely many roots and hence is an identity. Therefore, the desired statement holds if we take $a = 1$, $b = 0$ and $c = -1$.

Conclude, that it must be $ L(x)=(x^2-4)\left(P_n(x)^2 - 4\right) = \left(P_{n+1}(x) - P_{n-1}(x)\right)^2 =R(x)$. Then $P_n(s+s^{-1})=s^n+s^{-n}$ for all $s\in R\wedge s\neq 0\wedge n\in Z\setminus Z_-$ by induction. Hence for all $s\in R\setminus\lbrace 0\rbrace $ after some calculations $ L(s+s^{-1})-R(s+s^{-1})=0$, so polynomial $L(x)-R(x)$ is constant and equal $0$.

Why be clever when the same old tricks work? We claim that $\boxed{(a,b,c)=(1,0,-1)}$ is such a triplet. Consider the field of rational functions $\mathbb Q(x)$ and extend it by the roots of the polynomial $\nu^2-x\cdot\nu+ 1\in\mathbb Q(x)[\nu]$ to obtain the field $K\supseteq \mathbb Q(x)$; label these roots $r_1,r_2\in K$. It can be seen that $r_1\neq r_2$ (since otherwise we would have $(\nu^2-x\cdot\nu+ 1)'=2\nu-x=0\implies \nu=\tfrac{x}{2}$, a contradiction), and it follows from a well-known result that there exist $\text{``}$constants$\text{''}$ $\alpha_1,\alpha_2\in K$ such that for all $n$, $P_n$ may be expressed as $\alpha_1r_1^n+\alpha_2r_2^n$. Indeed, $\alpha_1=\alpha_2=1$ suffices (as can be seen from Viète's and the given values of $P$). The problem thus reduces to showing that $$(x^2 - 4)\left((r_1^n+r_2^n)^2- 4\right)=\left(r_1^n(r_1-r_1^{-1})+r_2^n(r_2-r_2^{-1})\right)^2$$Observe that for any root $r$ of $\nu^2-x\cdot\nu+ 1$ we have $r+r^{-1}=x$ and thus that $(r-r^{-1})^2=x^2-4$. In addition, $$(r_1-r_1^{-1})(r_2-r_2^{-1})=r_1r_2+r_1^{-1}r_2^{-1}-r_1r_2^{-1}-r_1^{-1}r_2=r_1r_2+\frac{1}{r_1r_2}-\frac{r_1^2+r_2^2}{r_1r_2}=4-x^2$$(Again by Viète's; this step may be replaced by the quadratic formula and computation). It it is thus sufficient to show that $$(x^2 - 4)\left((r_1^n+r_2^n)^2- 4\right)=r_1^{2n}(x^2-4)+r_2^{2n}(x^2-4)-2(x^2-4)\iff (r_1^n+r_2^n)^2-4=r_1^{2n}+r_2^{2n}-2$$Which is indeed the case $\blacksquare$