7. Find non-constant polynomials with integer coefficients $P(x)$ and $Q(x)$, satisfying
$$
P\left(Q^{2}(x)\right)=P(x) Q^{2}(x) \text {. }
$$
$$
\begin{array}{l}
\text { 7. } P(x)=a x^{2}, Q(x)=x ; \\
P(x)=a x^{2}, Q(x)=-x ; \\
P(x)=a x^{2}-4 a x, Q(x)=x-2 . \\
P(x)=a x^{2}-4 a x, Q(x)=-x+2 \text {. }
\end{array}
$$

Let the degrees of $P(x)$ and $Q(x)$ be $m$ and $n$ respectively.
Then the degree of $P\left(Q^{2}(x)\right)$ is $2 m n$, and the degree of $P(x) Q^{2}(x)$
is $m+2 n$.
$$
\begin{array}{l}
\text { Hence } 2 m n=m+2 n \\
\Rightarrow(m-1)(2 n-1)=1 \\
\Rightarrow m-1=2 n-1=1 \\
\Rightarrow m=2, n=1 .
\end{array}
$$

Let $P(x)=a x^{2}+b x+c$.
From the given information,
$$
a Q^{4}(x)+b Q^{2}(x)+c=P(x) Q^{2}(x) \text {. }
$$

Notice that the right side of equation (1) can be divided by $Q(x)$.
Thus, $c=0$.
Dividing both sides of equation (1) by $Q^{2}(x)$, we get
$$
P(x)=a Q^{2}(x)+b \text {. }
$$

Substituting the known information,
$$
\begin{array}{l}
a Q^{2}\left(Q^{2}(x)\right)+b=\left(a Q^{2}(x)+b\right) Q^{2}(x) \\
\Leftrightarrow a Q^{2}\left(Q^{2}(x)\right)=a Q^{4}(x)+b Q^{2}(x)-b .
\end{array}
$$

Since $Q(x)$ is not a constant, $y=Q^{2}(x)$ can take infinitely many values.
Thus, equation (2) can be expressed as
$$
\begin{array}{l}
a Q^{2}(y)=a y^{2}+b y-b \\
\Rightarrow Q^{2}(x)=x^{2}+\frac{b}{a} x-\frac{b}{a} .
\end{array}
$$

Since the coefficients of $Q(x)$ are integers, $\frac{b}{a}$ must be an integer.
Thus, $b=t a (t \in \mathbf{Z})$.
Hence, $Q^{2}(x)=x^{2}+t x-t$
$\Rightarrow x^{2}+t x-t$ is a perfect square
$\Rightarrow t^{2}+4 t=0$
$\Rightarrow t=0$ or $t=-4$.
If $t=0$, then $b=0, Q(x)=x$ and $Q(x)=-x$.
Thus, $P(x)=a x^{2}$ (where $a$ is a non-zero integer).
If $t=-4$, then $Q^{2}(x)=(x-2)^{2}$.
Thus, $Q(x)= \pm(x-2)$
$\Rightarrow P(x)=a(x-2)^{2}-4 a$ (where $a$ is a non-zero integer).
It is easy to verify that the above results satisfy the conditions of the problem.