Quidditch wrote: Find all polynomials $P\in\mathbb{Z}[x]$ such that $$|P(x)-x|\leq x^2+1$$for all real numbers $x$. Let $Q(x)=P(x)-x$ so that we want $|Q(x)|\le x^2+1$ $\forall x$ So $\lim_{x\to+\infty}\frac{Q(x)}{x^3}=0$ and $Q(x)$ has degree $\le 2$ So $Q(x)=ax^2+bx+c$ (with $a,b,c$ maybe zero) and we have $|ax^2+bx+c|\le x^2+1$ $\forall x$ $\iff$ $x^2+1\ge ax^2+bx+c$ and $x^2+1\ge -ax^2-bx-c$ $\forall x$ $\iff$ $(1-a)x^2-bx+1-c\ge 0$ and $(1+a)x^2+bx+1+c\ge 0$ $\forall x$ If $a=-1$, this is $2x^2-bx+1-c\ge 0$ and $bx+1+c\ge 0$ $\forall x$ Second condition implies $b=0$ and we have $2x^2+1-c\ge 0$ and $1+c\ge 0$ $\forall x$ $\iff$ $c\in[-1,+1]$ If $a=1$, this is $-bx+1-c\ge 0$ and $2x^2+bx+1+c\ge 0$ $\forall x$ First condition implies $b=0$ and we have $1-c\ge 0$ and $2x^2+1+c\ge 0$ $\forall x$ $\iff$ $c\in[-1,+1]$ If $a\ne\pm 1$, this is $1-a>0$ and $b^2-4(1-a)(1-c)\le 0$ and $1+a>0$ and $b^2-4(1+a)(1+c)\le 0$ $\iff$ $a\in(-1,1)$ and $c\in[-1,1]$ and $b^2\le 4(1+ac-|a+c|)$ (constraint $c\in[-1,1]$ is just to have at least one suitable $b$ ($1+ac-|a+c|\ge 0$) Hence the result : $P(x)=ax^2+(b+1)x+c$ with constraint : $|c|\le 1$ and : Either $|a|=1$ and then $b=0$ Either $|a|<1$ and $|b|\le 2\sqrt{1+ac-|a+c|}$

Quidditch wrote: Find all polynomials $P\in\mathbb{Z}[x]$ ... pco wrote: Hence the result : $P(x)=ax^2+(b+1)x+c$ with constraint : $|c|\le 1$ and : Either $|a|=1$ and then $b=0$ Either $|a|<1$ and $|b|\le 2\sqrt{1+ac-|a+c|}$ Uhhhh?

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