# Find all polynomials P(x) with real coefficents satisfying P^2(x)−1=4P(x^2−4x+1).

Find all polynomials $P\left(x\right)$ with real coefficents satisfying ${P}^{2}\left(x\right)-1=4P\left({x}^{2}-4x+1\right)$
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recepiamsb
Let $Q\left(x+3\right)=P\left(x\right)$. Hence
${Q}^{2}\left(x+3\right)-1=4Q\left({x}^{2}-4x+4\right)=4Q\left(\left(x-2{\right)}^{2}\right)$
Let $x\to x+2$
${Q}^{2}\left(x+5\right)=4Q\left({x}^{2}\right)+1$
So $Q\left(5+x\right)=±Q\left(5-x\right)$
Let $R\left(x\right)=Q\left(5+x\right)$ . R is either odd or even and
${R}^{2}\left(x\right)=4R\left({x}^{2}-5\right)+1$
If R is odd then $R\left(0\right)=0$ and $R\left(5\right)=\frac{1}{4}$ and $R\left(-20\right)=\frac{15}{64}$ and more generally :
${S}_{0}=0$, ${S}_{n+1}=5-{S}_{n}^{2}$, ${T}_{0}=0$ and ${T}_{n+1}=\frac{1-{T}_{n}^{2}}{4}$
It's easy to see that $P\left({S}_{n}\right)={T}_{n}$, but $lim\left({T}_{n}\right)$ is finite and $lim\left(|{S}_{n}|\right)=\mathrm{\infty }$ is not a polynomial function.
If R is even, $R\left(0\right)$ is an optimum (either maximum or minimum, ${R}^{\prime }\left(0\right)=0$). Hence $R\left(\sqrt{5}\right)$ is an optimum too, and ${S}_{0}=0$, ${S}_{n+1}=\sqrt{5+{S}_{n}}$ , then ${S}_{n}$ are all optimum... But all ${S}_{n}$ are different, so ${R}^{\prime }$ has an infinite number of roots. This is not a polynomial function, except for a degree 0 polynomial.