4. Given real numbers $a, b, x, y$ satisfy
$$
\begin{array}{l}
a x+b y=3, a x^{2}+b y^{2}=7, \\
a x^{3}+b y^{3}=16, a x^{4}+b y^{4}=42 .
\end{array}
$$

Then $a x^{5}+b y^{5}=$ $\qquad$
4.20 .
$$
\begin{array}{l}
\text { Given } a x^{3}+b y^{3}=16 \\
\Rightarrow\left(a x^{3}+b y^{3}\right)(x+y)=16(x+y) \\
\Rightarrow\left(a x^{4}+b y^{4}\right)+x y\left(a x^{2}+b y^{2}\right)=16(x+y) \\
\Rightarrow 42+7 x y=16(x+y), \\
a x^{2}+b y^{2}=7 \\
\Rightarrow\left(a x^{2}+b y^{2}\right)(x+y)=7(x+y) \\
\Rightarrow\left(a x^{3}+b y^{3}\right)+x y(a x+b y)=7(x+y) \\
\Rightarrow 16+3 x y=7(x+y),
\end{array}
$$

Solving equations (1) and (2) simultaneously, we get $x+y=-14, x y=-38$.
$$
\begin{array}{l}
\text { Then } a x^{4}+b y^{4}=42 \\
\Rightarrow\left(a x^{4}+b y^{4}\right)(x+y)=42(x+y) \\
\Rightarrow\left(a x^{5}+b y^{5}\right)+x y\left(a x^{3}+b y^{3}\right)=42(x+y) \\
\Rightarrow a x^{5}+b y^{5}=42(x+y)-16 x y=20 .
\end{array}
$$