A sequence consists of positive numbers, and for any three consecutive elements, the middle number is the product of the two outer numbers. The product of the first five elements, as well as the product of the next five elements, is 2. Determine the first ten terms of the sequence.
I. solution. Let's write down the product of the first five elements of the sequence, using the relationship between the elements:

$$
a_{1} \cdot a_{2} \cdot \frac{a_{2}}{a_{1}} \cdot \frac{a_{3}}{a_{2}} \cdot \frac{a_{4}}{a_{3}}=a_{2} \cdot a_{4}=2
$$

By substituting $a_{4}=\frac{a_{3}}{a_{2}}$, we get $a_{2} \frac{a_{3}}{a_{2}}=a_{3}=2$.

Similarly, from the product of the next five elements, we get $a_{8}=2$.

Then, let's write down the product of the first ten elements in the following form:

$$
a_{1} \cdot a_{2} \cdot \frac{a_{2}}{a_{1}} \cdot \frac{a_{3}}{a_{2}} \cdot \frac{a_{4}}{a_{3}} \cdot \frac{a_{5}}{a_{4}} \cdot \frac{a_{6}}{a_{5}} \cdot \frac{a_{7}}{a_{6}} \cdot \frac{a_{8}}{a_{7}} \cdot \frac{a_{9}}{a_{8}}=4
$$

After simplifying, using the condition, and substituting the previously found values of $a_{3}$ and $a_{8}$:

$$
a_{2} \cdot a_{9}=a_{3} \cdot a_{1} \cdot a_{8} \cdot a_{10}=4 \cdot a_{1} \cdot a_{10}=4
$$

From this, $a_{1}=\frac{1}{a_{10}}, a_{2}=a_{1} \cdot a_{3}=\frac{1}{a_{10}} \cdot 2=\frac{2}{a_{10}}, a_{4}=\frac{a_{3}}{a_{2}}=\frac{2}{\frac{2}{a_{10}}}=a_{10}, a_{5}=\frac{a_{4}}{a_{3}}=\frac{a_{10}}{2}, a_{6}=\frac{a_{5}}{a_{4}}=\frac{a_{10}}{2 a_{10}}=\frac{1}{2}$, $a_{7}=a_{6} \cdot a_{8}=\frac{1}{2} \cdot 2=1, a_{9}=2, a_{10}=1$, and finally $a_{5}=\frac{1}{2}, a_{4}=1, a_{2}=2$ and $a_{1}=1$.

The elements of the sequence in order: $1,2,2,1, \frac{1}{2}, \frac{1}{2}, 1,2,2,1$. It is easy to verify that the obtained sequence satisfies the condition.

Szeberényi Ágnes (Budapest, Árpád Gym., 10th grade)

II. solution. Our numbers are positive, so let's consider their base-2 logarithms, let $b_{i}=\log _{2} a_{i}$. The condition for them means that

$$
b_{i}=b_{i-1}+b_{i+1} \quad i=2,3, \ldots, 9
$$

and $b_{1}+b_{2}+b_{3}+b_{4}+b_{5}=b_{6}+b_{7}+b_{8}+b_{9}+b_{10}=1$. Add (1) for $i$ and $(i+1)$:

$$
b_{i}+b_{i+1}=b_{i-1}+b_{i+1}+b_{i}+b_{i+2}, \quad \text { hence } \quad 0=b_{i-1}+b_{i+2}
$$

the "third-neighbors" of the $b_{i}$ sequence are therefore each other's opposites. If $b_{1}=x, b_{2}=y$ and $b_{3}=z$, then the elements of the sequence are respectively

| $b_{1}$ | $b_{2}$ | $b_{3}$ | $b_{4}$ | $b_{5}$ | $b_{6}$ | $b_{7}$ | $b_{8}$ | $b_{9}$ | $b_{10}$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $x$ | $y$ | $z$ | $-x$ | $-y$ | $-z$ | $x$ | $y$ | $z$ | $-x$ |

Thus $b_{1}+b_{2}+b_{3}+b_{4}+b_{5}=z=1$ and $b_{6}+b_{7}+b_{8}+b_{9}+b_{10}=y=1$, so $y=z=1$. Since (1) states $y=x+z$, from this $x=0$. Therefore, the $b_{i}$ sequence is $0,1,1,0,-1,-1,0,1,1,0$, and the $a_{i}$ sequence accordingly $a_{i}=2^{b_{i}}$ is $1,2,2,1, \frac{1}{2}, \frac{1}{2}, 1,2,2,1$.

From the above, it follows that if the problem has a solution, then it can only be the found sequence. However, if the condition is satisfied for the $(x ; y ; z)$ triplet, that is, $x-y+z=0$, then we get the $(y ; z ;-x)$ triplet by shifting this triplet by one in the $b_{i}$ sequence. Calculating the sum $y-z+(-x)$, we get $y-z-x=-(x-y+z)=0$ as well.

If, therefore, the second element is the sum of the first and the third, then according to the rule of forming the $b_{i}$ sequence, every element is the sum of the two surrounding it, so our sequence is the only solution to the problem.

Remark. This line of reasoning would be much less comprehensible in the language of products and quotients, but the transition to logarithms not only simplifies the solution typographically. By writing the condition in the form (1) and rearranging, we get the linear recurrence $b_{i+1}=b_{i}-b_{i-1}$, and there are well-known results for sequences with such properties.