Slovenija 2008

Since $a_1,a_2,a_3,a_4,a_5$ are terms of a geometric sequence, we can write $a_i=a_1\cdot q^{i-1}$ for $i=2,3,4,5$ and for some real number $q$. But $q=\frac{a_2}{a_1}$ is a quotient of two positive integers, so $q$ is rational. Write $q=\frac{m}{n}$ as a reduced fraction. The terms of the sequence are $$a_1,\frac{a_1\cdot m}{n},\frac{a_1\cdot m^2}{n^2},\frac{a_1\cdot m^3}{n^3},\frac{a_1\cdot m^4}{n^4}.$$ Since all terms are positive integers and $m$ and $n$ are relatively prime, we see that $n^4$ is a divisor of $a_1$. So, we can write $a_1=d{n}^4$, where $d$ is some positive integer. Therefore, the terms of the sequence are $$d{n}^4,dm{n}^3,d{m}^2{n}^2,d{m}^{3}n,d{m}^4.$$ But since no prime number divides all five of these terms, we have $d=1$, and the sequence becomes $n^4,m{n}^3,m^2{n}^2,m^{3}n,m^4$. We also know that the terms are all less than $2008$, so $m^4<2008$ and $n^4<2008$. This implies $m\le 6$ and $n\le 6$. But $a_1=n^4$ is not divisible by $6$, so $n\le 5$. We also know that $a_2=n^{3}m$ is divisible by $5$, $a_3=n^2{m}^2$ is divisible by $4$ and $a_4=m^{3}n$ is divisible by $3$, so the product $mn$ is divisible by $2$, $3$ and $5$, i.e. by $30$. At the same time $mn\le 6\cdot 5=30$, so this product is equal to $30$. This implies $m=6$ and $n=5$. The terms of the sequence are $5^4=625$, $5^3\cdot 6=750$, $5^2\cdot 6^2=900$, $5\cdot 6^3=1080$ and $6^4=1296$.