3. All members of an infinite geometric progression are natural numbers. The sum of the third, fifth, and seventh terms of this progression is equal to $819 \cdot 6^{2016}$. Find the common ratio of the progression. (8 points)
Solution: We have a geometric progression $b_{1}, b_{1} q, b_{1} q^{2}, \cdots, b_{1} q^{n-1}, \cdots$, and $b_{1} q^{n-1} \in N$ for any number $n \in N$. Thus, $b_{1}$ and $q$ are natural numbers. According to the condition $b_{3}+b_{5}+b_{7}=819 \cdot 6^{2016}$, or $b_{1} q^{2}+b_{1} q^{4}+b_{1} q^{6}=2^{2016} \cdot 3^{2018} \cdot 7 \cdot 13$, $b_{1} q^{2}\left(1+q^{2}+q^{4}\right)=2^{2016} \cdot 3^{2018} \cdot 7 \cdot 13$. The natural number $1+q^{2}+q^{4}$ for any $q \in N$ is an odd number, therefore, $1+q^{2}+q^{4}=3^{k} \cdot 7^{l} \cdot 13^{m}$, where $k \in\{0,1, \ldots, 2018\}, \quad l, m \in\{0,1\}$.

1) If $k=0$, then $1+q^{2}+q^{4}=7^{l} \cdot 13^{m}, l \in\{0,1\}$.
a) When $l=0$ we get the equation $q^{2}+q^{4}=13^{m}$, which has no natural solutions (the discriminant $D=1+4 \cdot 13^{m}$ equals 5 when $m=0$, and equals 53 when $m=1$).

b) When $l=1, m=0$ we get the equation $q^{2}+q^{4}-7=0$, which has no natural solutions ( $D=29$).

c) When $l=1, m=1$ we get the equation $q^{2}+q^{4}-90=0, q^{2}=9, q=3$. In this case, $b_{1}=6^{2016}$.

2) If $k=1$, then $1+q^{2}+q^{4}=3 \cdot 7^{l} \cdot 13^{m}, l, m \in\{0,1\}$.

a) When $l=0, m=0$ we get the equation $q^{2}+q^{4}-2=0, q^{2}=1, q=1$. In this case, $b_{1}=6^{2016} \cdot 3 \cdot 91$.
b) When $l=0, m=1$ we get the equation $q^{2}+q^{4}-38=0$, which has no natural solutions ( $D=153$ ).

c) When $l=1, m=0$ we get the equation $q^{2}+q^{4}-20=0, q^{2}=4, q=2$. In this case, $b_{1}=2^{2014} \cdot 3^{2017} \cdot 13$.

d) When $l=1, m=1$ we get the equation $q^{2}+q^{4}-272=0, q^{2}=16, q=4$. In this case, $b_{1}=2^{2012} \cdot 3^{2017} \cdot$

3) If $k \in\{2, \ldots, 2018\}, l, m \in\{0,1\}$, then $1+q^{2}+q^{4}=3^{k} \cdot 7^{l} \cdot 13^{m}$. For the obtained biquadratic equation $q^{2}+q^{4}+1-3^{k} \cdot 7^{l} \cdot 13^{m}=0$ we compute the discriminant $D=1-4\left(1-3^{k} \cdot 7^{l} \cdot 13^{m}\right)=3\left(4 \cdot 3^{k-1} \cdot 7^{l} \cdot 13^{m}-1\right)$. Since for $k \in\{2, \ldots, 2018\}, l, m \in\{0,1\}$, the number $4 \cdot 3^{k-1} \cdot 7^{l} \cdot 13^{m}$ is divisible by 3, then $4 \cdot 3^{k-1} \cdot 7^{l} \cdot 13^{m}-1$ is not divisible by 3, and $\sqrt{D}$ is an irrational number. Therefore, the equation $q^{2}+q^{4}+1-3^{k} \cdot 7^{l} \cdot 13^{m}=0$ has no natural roots.

Answer: The problem has four solutions: 1) $q=1$; 2) $q=2$; 3) $q=3$; 4) $q=4$.