(IMO SL 2014 A5) Consider all polynomials $P$ with real coefficients having the following property:

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

Determine all possible values of $P(0)$.
Answer: $P(0) \in]-\infty, 0[\cup\{1\}$.

The lengths of the proof are only technical, but it can be summarized in the following steps:

Step 1: $C \leqslant 0$ works: Take a simple polynomial $P$ of degree 2.

Step 2: If $P(0) \geqslant 0$, then $P$ is even: look at $(x, y)$ and $(-x, y)$ and look at $x$ very small + rigidity.

Step 3: $P>0$ Look at $x$ close to a root and for $y=0$.

Step 4: $P$ is of degree $2$: look at $x$ large and compare degrees.

Step 5: $P(x)=x^{2}+1$: calculations.

Step 6: $P(0)=1$ works, do inequalities.

Step 1: First, if $-C$ is negative, we look for a simple polynomial such that $P(0)=-C$ and the inequality $\left|y^{2}-P(x)\right|>2|x|$ for all $x, y$. We can take a polynomial that is strictly positive so that the absolute value disappears. The mean value theorem suggests choosing $P(x)=-\left((1+\varepsilon) \frac{x^{2}}{C}+C\right)$. Then we have

$$
\left|y^{2}-P(x)\right|=y^{2}+\left((1+\varepsilon) \frac{x^{2}}{C}+C\right) \geqslant \varepsilon \frac{x^{2}}{C}+2|x|
$$

with equality if and only if $x=C$, so that $|y-P(x)|>2|x|$ and $P$ satisfies the equivalence.

## Step 2:

We now assume that $P(0) \geqslant 0$.

We show that $P$ is even. For this, we apply the hypothesis to the pairs $(x, y)$ and $(-x, y)$ to find

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

We deduce in particular that

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

This equivalence is interesting only if $P(x)+2|x| \geqslant 0$ for an infinite number of values of $x$. For this, we look close to 0. The reasoning is: close to 0, we have for $P(x)+$ $2|x| \sim P(0)+P^{\prime}(0) x+2|x|$. So if $P^{\prime}(0) \geqslant 2$, we find $P(x) \geqslant 0$ around 0, and if $P^{\prime}(0) \leqslant 2$, we find $P(-x) \geqslant 0$ around 0.

The rigorous version is: we have $M>0$ such that $\left|P(x)-\left(P(0)+x P^{\prime}(0)\right)\right| \leqslant M x^{2}$ (since $P^{\prime}(0)$ corresponds to the coefficient in front of $x$ of $\left.P\right)$. Thus, we have

$$
P(x)=P(x)-\left(P(0)+x P^{\prime}(0)\right)+\left(P(0)+x P^{\prime}(0)\right)
$$

For $x$ sufficiently small, we have $\left|P(x)-\left(P(0)+x P^{\prime}(0)\right)\right|0$ for $x$ non-zero so that $\left|x^{2} Q(x)\right|>2|x|$ which is absurd for $x$ large. We deduce that $P(0)>0$.

Step 4: $P$ is quadratic.

We assume $\operatorname{deg} P \geqslant 4$. By taking $y=\sqrt{P(x)}$, the hypothesis is not empty and we get $P(\sqrt{P(x)}) \leqslant x^{2}+2 \sqrt{P(x)}$. Since $P$ is strictly positive and diverges to $+\infty$, it tends to $+\infty$ as $x \to +\infty$. But by comparing degrees, we get a contradiction for $x$ sufficiently large (the right side of the inequality grows faster than the left side).

Step 5: $P(x)=x^{2}+1$.

We write $P(x)=a x^{2}+b$. By taking $y=\sqrt{a} x$, for $x$ large, the equivalence is not empty and gives that $\left|\left(1-a^{2}\right) x^{2}-b\right| \leqslant 2 \sqrt{a} x$, which forces $a=1$. To get $b$, we replace $y$ with $x+1$ to obtain the equivalence

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

which gives for $x$ sufficiently large that $b \in[1,4 x+1] \Longleftrightarrow b \in[-4 x-3,1]$. We deduce that $b=1$ by taking $x$ such that $b \in[-4 x-3,4 x+1]$.

Thus, $P(x)=x^{2}+1$.

Step 6: Conversely, $P(x)=x^{2}+1$ satisfies the equivalence.

We verify that

$$
\left(y^{2}-x^{2}-1\right)^{2} \leqslant 4 x^{2} \Longleftrightarrow\left((x+y)^{2}-1\right)\left(1-(x-y)^{2}\right) \Longleftrightarrow\left(x^{2}-y^{2}-1\right)^{2} \leqslant 4 y^{2}
$$