7.1. Given two different geometric progressions, the first terms of which are equal to 1, and the sum of the denominators is 3. Find the sum of the fifth terms of these progressions, if the sum of the sixth terms is 573. If the answer to the question is not unique, specify the sum of all possible values of the desired quantity.
Answer: 161. Solution. Let the denominators of the progressions be $p$ and $q$. According to the condition, we get: $\left\{\begin{array}{c}p+q=3, \\ p^{5}+q^{5}=573\end{array}\right.$. We need to find $p^{4}+q^{4}$. Denoting $p+q=a, p q=b$, we express:

$$
\begin{gathered}
p^{4}+q^{4}=\left((p+q)^{2}-2 p q\right)^{2}-2 p^{2} q^{2}=\left(a^{2}-2 b\right)^{2}-2 b^{2}=a^{4}-4 a^{2} b+2 b^{2} ; \\
p^{5}+q^{5}=(p+q)\left(p^{4}-p^{3} q+p^{2} q^{2}-p q^{3}+q^{4}\right)=(p+q)\left(p^{4}+q^{4}-p q\left(p^{2}+q^{2}\right)+p^{2} q^{2}\right) \\
=a\left(a^{4}-4 a^{2} b+2 b^{2}-b\left(a^{2}-2 b\right)+b^{2}\right)=a\left(a^{4}-5 a^{2} b+5 b^{2}\right) .
\end{gathered}
$$

$$
\text { According to the condition }\left\{\begin{array} { c } 
{ a = 3 , } \\
{ a ( a ^ { 4 } - 5 a ^ { 2 } b + 5 b ^ { 2 } ) = 5 7 3 }
\end{array} \Leftrightarrow \left\{\begin{array} { c } 
{ a = 3 , } \\
{ 8 1 - 4 5 b + 5 b ^ { 2 } = 1 9 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{c}
a=3, \\
b^{2}-9 b-22=0
\end{array} .\right.\right.\right.
$$

From here, $b=11$ or $b=-2$. Since the system $\left\{\begin{array}{c}p+q=3, \\ p q=11\end{array}\right.$ has no solutions, then $b=-2$.

Therefore, $a=3$ and $b=-2$. Thus, $p^{4}+q^{4}=a^{4}-4 a^{2} b+2 b^{2}=81+72+8=161$.