5. Consider all real-coefficient polynomials $P(x)$ that satisfy the following property: for any two real numbers $x, y$, we have
$$
\begin{array}{l}
\left|y^{2}-P(x)\right| \leqslant 2|x| \\
\Leftrightarrow\left|x^{2}-P(y)\right| \leqslant 2|y| .
\end{array}
$$

Find all possible values of $P(0)$.
5. The set of all possible values of $P(0)$ is $(-\infty, 0) \cup\{1\}$.
(1) First, prove that these values of $P(0)$ can be achieved.

For each negative real number $-C$, the polynomial
$$
P(x)=-\left(\frac{2 x^{2}}{C}+C\right)
$$

satisfies $P(0)=-C$.
To prove that $P(x)$ satisfies the required properties in the problem, by symmetry, it suffices to prove that for each $C>0$ and any two real numbers $x, y$, we have
$$
\left|y^{2}-P(x)\right|>2|x| \text {. }
$$

In fact,
$$
\begin{array}{l}
\left|y^{2}-P(x)\right|=y^{2}+\frac{x^{2}}{C}+\frac{(|x|-C)^{2}}{C}+2|x| \\
\geqslant \frac{x^{2}}{C}+2|x| \geqslant 2|x| .
\end{array}
$$

Note that the first inequality holds only when $|x|=C$, and the second inequality holds only when $x=0$. Since these two conditions cannot hold simultaneously, then
$$
\begin{array}{l}
\left|y^{2}-P(x)\right|>2|x| \text {. } \\
\text { The polynomial } P(x)=x^{2}+1 \text { satisfies } P(0)=1 . \\
\text { Now prove: } P(x) \text { satisfies (1) and (2). } \\
\text { Note that, for all real numbers } x, y \text {, we have } \\
\left|y^{2}-P(x)\right| \leqslant 2|x| \\
\Leftrightarrow\left(y^{2}-x^{2}-1\right)^{2} \leqslant 4 x^{2} \\
\Leftrightarrow 0 \leqslant\left[y^{2}-(x-1)^{2}\right]\left[(x+1)^{2}-y^{2}\right] \\
\Leftrightarrow 0 \leqslant(y-x+1)(y+x-1)(x+1-y)(x+1+y) \\
\Leftrightarrow 0 \leqslant\left[(x+y)^{2}-1\right]\left[1-(x-y)^{2}\right] .
\end{array}
$$

Since the above inequality is symmetric in $x, y$, it is also equivalent to
$$
\left|x^{2}-P(y)\right| \leqslant 2|y| \text {. }
$$
(2) Next, prove that $P(0)$ cannot take any other values.

To prove this, assume $P(x)$ satisfies (1), and $P(0) \geqslant 0$.
We will prove in four steps that for all real numbers $x$, we have
$$
P(x)=x^{2}+1 \Rightarrow P(0)=1 \text {. }
$$

【Step 1】 $P(x)$ is an even function.
From (1) and (2), we know that for any real numbers $x, y$,
$$
\left|y^{2}-P(x)\right| \leqslant 2|x|
$$
$\Leftrightarrow\left|x^{2}-P(y)\right| \leqslant 2|y|$
$$
\Leftrightarrow\left|y^{2}-P(-x)\right| \leqslant 2|x| \text {. }
$$

Considering that the first inequality is equivalent to the third inequality and $y^{2}$ takes values in $\mathbf{R} \backslash \mathbf{R}_{-}$, then for all $x \in \mathbf{R}$, we have $[P(x)-2|x|, P(x)+2|x|] \cap \mathbf{R} \backslash \mathbf{R}_{-}$ $=[P(-x)-2|x|, P(-x)+2|x|] \cap \mathbf{R} \backslash \mathbf{R}_{-}$. We will prove: there are infinitely many real numbers $x$ such that
$$
P(x)+2|x| \geqslant 0 \text {. }
$$

In fact, for any real-coefficient polynomial satisfying $P(0) \geqslant 0$, the conclusion holds. For this, assume the coefficient of $x$ in $P(x)$ is non-negative. In this case, for sufficiently small positive real numbers $x$, the conclusion holds.

For the real numbers $x$ satisfying $P(x)+2|x| \geqslant 0$, we have
$$
\begin{array}{l}
P(x)+2|x|=P(-x)+2|x| \\
\Rightarrow P(x)=P(-x) .
\end{array}
$$

Since the polynomial $P(x)-P(-x)$ has infinitely many zeros, it is identically zero, i.e., $P(x)=P(-x)$ for all real numbers $x$.
Therefore, $P(x)$ is an even function.
【Step 2】 For all $t \in \mathbf{R}$, $P(t)>0$.
Assume there exists a real number $t \neq 0$ such that $P(t)=0$. Then there exists an open interval $I$ such that for all $y \in I$, we have
$|P(y)| \leqslant 2|y|$.
Taking $x=0$ in (1), then for all $y \in I$ we have $y^{2}=P(0)$, which is a contradiction.
$\left|\left(1-a^{2}\right) x^{2}-b\right| \leqslant 2 \sqrt{a} x$.
By $a>0$, we know that it only holds when $a=1$.
Taking $y=x+1(x>0)$ in (1), then
$|2 x+1-b| \leqslant 2 x$
$$
\Leftrightarrow|2 x+1+b| \leqslant 2 x+2 \text {, }
$$

i.e., for all $x>0$,
$b \in[1,4 x+1] \Leftrightarrow b \in[-4 x-3,1]$. This is only possible when $b=1$.
Thus, $P(x)=x^{2}+1$.