4. The largest integer not exceeding $(\sqrt{5}+\sqrt{3})^{6}$ is
$\qquad$
4.3903.

Notice that $(\sqrt{5}+\sqrt{3})^{6}=(8+2 \sqrt{15})^{3}$.
Let $8+2 \sqrt{15}=a, 8-2 \sqrt{15}=b$. Then we have
$$
a+b=16, ab=4 \text{. }
$$

We know that $a$ and $b$ are the roots of the equation $x^{2}-16 x+4=0$, so we have
$$
\begin{array}{l}
a^{2}=16 a-4, b^{2}=16 b-4 ; \\
a^{3}=16 a^{2}-4 a, b^{3}=16 b^{2}-4 b . \\
\text{Therefore, } a^{3}+b^{3}=16\left(a^{2}+b^{2}\right)-4(a+b) \\
=16[16(a+b)-8]-4(a+b) \\
=252(a+b)-128=3904 .
\end{array}
$$

Since $0<b<1$, it follows that $3903<a^{3}<3904$.
Therefore, the largest integer not exceeding $(\sqrt{5}+\sqrt{3})^{6}$ is 3903.