4.056. Find the fifth term of an increasing geometric progression, given that its first term is equal to $7-3 \sqrt{5}$ and that each of its terms, starting from the second, is equal to the difference of the two adjacent terms.
Solution.

We have:

$$
\left\{\begin{array}{l}
b_{1}=7-3 \sqrt{5} \\
b_{2}=b_{3}-b_{1}, \\
|q|>1
\end{array}\right.
$$

$q_{1}=\frac{1-\sqrt{5}}{2}$ does not fit, as $\left|q_{1}\right|<1 ; q_{2}=\frac{1+\sqrt{5}}{2}$.

From this,

$$
b_{5}=b_{1} q^{4}=\frac{(7-3 \sqrt{5})(1+\sqrt{5})^{4}}{16}=\frac{(17-3 \sqrt{5}) 8(7+3 \sqrt{5})}{16}=\frac{49-45}{2}=2
$$

Answer: 2.