Subfield: Algebra


Question: Consider all polynomials $P(x)$ with real coefficients that have the following property: for any two real numbers $x$ and $y$ one has

$$
\left|y^{2}-P(x)\right| \leqslant 2|x| \text { if and only if }\left|x^{2}-P(y)\right| \leqslant 2|y|
\tag{1}
$$

Determine all possible values of $P(0)$.

Solution:
Part I. We begin by verifying that these numbers are indeed possible values of $P(0)$. To see that each negative real number $-C$ can be $P(0)$, it suffices to check that for every $C>0$ the polynomial $P(x)=-\left(\frac{2 x^{2}}{C}+C\right)$ has the property described in the statement of the problem. Due to symmetry it is enough for this purpose to prove $\left|y^{2}-P(x)\right|>2|x|$ for any two real numbers $x$ and $y$. In fact we have

$$
\left|y^{2}-P(x)\right|=y^{2}+\frac{x^{2}}{C}+\frac{(|x|-C)^{2}}{C}+2|x| \geqslant \frac{x^{2}}{C}+2|x| \geqslant 2|x|
$$

where in the first estimate equality can only hold if $|x|=C$, whilst in the second one it can only hold if $x=0$. As these two conditions cannot be met at the same time, we have indeed $\left|y^{2}-P(x)\right|>2|x|$.

To show that $P(0)=1$ is possible as well, we verify that the polynomial $P(x)=x^{2}+1$ satisfies (1). Notice that for all real numbers $x$ and $y$ we have

$$
\begin{aligned}
\left|y^{2}-P(x)\right| \leqslant 2|x| & \Longleftrightarrow\left(y^{2}-x^{2}-1\right)^{2} \leqslant 4 x^{2} \\
& \Longleftrightarrow 0 \leqslant\left(\left(y^{2}-(x-1)^{2}\right)\left((x+1)^{2}-y^{2}\right)\right. \\
& \Longleftrightarrow 0 \leqslant(y-x+1)(y+x-1)(x+1-y)(x+1+y) \\
& \Longleftrightarrow 0 \leqslant\left((x+y)^{2}-1\right)\left(1-(x-y)^{2}\right) .
\end{aligned}
$$

Since this inequality is symmetric in $x$ and $y$, we are done.

Part II. Now we show that no values other than those mentioned in the answer are possible for $P(0)$. To reach this we let $P$ denote any polynomial satisfying (1) and $P(0) \geqslant 0$; as we shall see, this implies $P(x)=x^{2}+1$ for all real $x$, which is actually more than what we want.

First step: We prove that $P$ is even.

By (1) we have

$$
\left|y^{2}-P(x)\right| \leqslant 2|x| \Longleftrightarrow\left|x^{2}-P(y)\right| \leqslant 2|y| \Longleftrightarrow\left|y^{2}-P(-x)\right| \leqslant 2|x|
$$

for all real numbers $x$ and $y$. Considering just the equivalence of the first and third statement and taking into account that $y^{2}$ may vary through $\mathbb{R}_{\geqslant 0}$ we infer that

$$
[P(x)-2|x|, P(x)+2|x|] \cap \mathbb{R}_{\geqslant 0}=[P(-x)-2|x|, P(-x)+2|x|] \cap \mathbb{R}_{\geqslant 0}
$$

holds for all $x \in \mathbb{R}$. We claim that there are infinitely many real numbers $x$ such that $P(x)+2|x| \geqslant 0$. This holds in fact for any real polynomial with $P(0) \geqslant 0$; in order to see this, we may assume that the coefficient of $P$ appearing in front of $x$ is nonnegative. In this case the desired inequality holds for all sufficiently small positive real numbers.

For such numbers $x$ satisfying $P(x)+2|x| \geqslant 0$ we have $P(x)+2|x|=P(-x)+2|x|$ by the previous displayed formula, and hence also $P(x)=P(-x)$. Consequently the polynomial $P(x)-P(-x)$ has infinitely many zeros, wherefore it has to vanish identically. Thus $P$ is indeed even.



Second step: We prove that $P(t)>0$ for all $t \in \mathbb{R}$.

Let us assume for a moment that there exists a real number $t \neq 0$ with $P(t)=0$. Then there is some open interval $I$ around $t$ such that $|P(y)| \leqslant 2|y|$ holds for all $y \in I$. Plugging $x=0$ into (1) we learn that $y^{2}=P(0)$ holds for all $y \in I$, which is clearly absurd. We have thus shown $P(t) \neq 0$ for all $t \neq 0$.

In combination with $P(0) \geqslant 0$ this informs us that our claim could only fail if $P(0)=0$. In this case there is by our first step a polynomial $Q(x)$ such that $P(x)=x^{2} Q(x)$. Applying (1) to $x=0$ and an arbitrary $y \neq 0$ we get $|y Q(y)|>2$, which is surely false when $y$ is sufficiently small.

Third step: We prove that $P$ is a quadratic polynomial.

Notice that $P$ cannot be constant, for otherwise if $x=\sqrt{P(0)}$ and $y$ is sufficiently large, the first part of (1) is false whilst the second part is true. So the degree $n$ of $P$ has to be at least 1 . By our first step $n$ has to be even as well, whence in particular $n \geqslant 2$.

Now assume that $n \geqslant 4$. Plugging $y=\sqrt{P(x)}$ into (1) we get $\left|x^{2}-P(\sqrt{P(x)})\right| \leqslant 2 \sqrt{P(x)}$ and hence

$$
P(\sqrt{P(x)}) \leqslant x^{2}+2 \sqrt{P(x)}
$$

for all real $x$. Choose positive real numbers $x_{0}, a$, and $b$ such that if $x \in\left(x_{0}, \infty\right)$, then $a x^{n}<$ $P(x)<b x^{n}$; this is indeed possible, for if $d>0$ denotes the leading coefficient of $P$, then $\lim _{x \rightarrow \infty} \frac{P(x)}{x^{n}}=d$, whence for instance the numbers $a=\frac{d}{2}$ and $b=2 d$ work provided that $x_{0}$ is chosen large enough.

Now for all sufficiently large real numbers $x$ we have

$$
a^{n / 2+1} x^{n^{2} / 2}<a P(x)^{n / 2}<P(\sqrt{P(x)}) \leqslant x^{2}+2 \sqrt{P(x)}<x^{n / 2}+2 b^{1 / 2} x^{n / 2},
$$

i.e.

$$
x^{\left(n^{2}-n\right) / 2}<\frac{1+2 b^{1 / 2}}{a^{n / 2+1}}
$$

which is surely absurd. Thus $P$ is indeed a quadratic polynomial.

Fourth step: We prove that $P(x)=x^{2}+1$.

In the light of our first three steps there are two real numbers $a>0$ and $b$ such that $P(x)=$ $a x^{2}+b$. Now if $x$ is large enough and $y=\sqrt{a} x$, the left part of (1) holds and the right part reads $\left|\left(1-a^{2}\right) x^{2}-b\right| \leqslant 2 \sqrt{a} x$. In view of the fact that $a>0$ this is only possible if $a=1$. Finally, substituting $y=x+1$ with $x>0$ into (1) we get

$$
|2 x+1-b| \leqslant 2 x \Longleftrightarrow|2 x+1+b| \leqslant 2 x+2,
$$

i.e.,

$$
b \in[1,4 x+1] \Longleftrightarrow b \in[-4 x-3,1]
$$

for all $x>0$. Choosing $x$ large enough, we can achieve that at least one of these two statements holds; then both hold, which is only possible if $b=1$, as desired.

Final answer: $(-\infty, 0) \cup\{1\}$.