vagnhestagn

2022-10-08

Find all polynomials $P(x)$ with real coefficents satisfying ${P}^{2}(x)-1=4P({x}^{2}-4x+1)$

recepiamsb

Beginner2022-10-09Added 9 answers

Let $Q(x+3)=P(x)$. Hence

${Q}^{2}(x+3)-1=4Q({x}^{2}-4x+4)=4Q((x-2{)}^{2})$

Let $x\to x+2$

${Q}^{2}(x+5)=4Q({x}^{2})+1$

So $Q(5+x)=\pm Q(5-x)$

Let $R(x)=Q(5+x)$ . R is either odd or even and

${R}^{2}(x)=4R({x}^{2}-5)+1$

If R is odd then $R(0)=0$ and $R(5)=\frac{1}{4}$ and $R(-20)=\frac{15}{64}$ 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 $P({S}_{n})={T}_{n}$, but $lim({T}_{n})$ is finite and $lim(|{S}_{n}|)=\mathrm{\infty}$ is not a polynomial function.

If R is even, $R(0)$ is an optimum (either maximum or minimum, ${R}^{\prime}(0)=0$). Hence $R(\sqrt{5})$ 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.

${Q}^{2}(x+3)-1=4Q({x}^{2}-4x+4)=4Q((x-2{)}^{2})$

Let $x\to x+2$

${Q}^{2}(x+5)=4Q({x}^{2})+1$

So $Q(5+x)=\pm Q(5-x)$

Let $R(x)=Q(5+x)$ . R is either odd or even and

${R}^{2}(x)=4R({x}^{2}-5)+1$

If R is odd then $R(0)=0$ and $R(5)=\frac{1}{4}$ and $R(-20)=\frac{15}{64}$ 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 $P({S}_{n})={T}_{n}$, but $lim({T}_{n})$ is finite and $lim(|{S}_{n}|)=\mathrm{\infty}$ is not a polynomial function.

If R is even, $R(0)$ is an optimum (either maximum or minimum, ${R}^{\prime}(0)=0$). Hence $R(\sqrt{5})$ 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.

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