72nd Czech and Slovak Mathematical Olympiad

The first equation of the given system is fulfilled if and only if the following two conditions are satisfied: ▷ the number $7z$ is integer, ▷ $7z\le 3x+5y+7z<7z+1$, i.e. $3x+5y\in [0,1)$.

Similarly, the second and third equations are fulfilled if and only if the numbers $7x$ and $7y$ are integers and $3y+5z$, $3z+5x\in [0,1)$.

Now consider any triple of non-negative numbers $(x,y,z)$, which is the solution to the problem. The inequalities $z\ge 0$ and $3z+5x<1$ imply $5x<1$, whence $7x<7/5<2$. This means that non-negative integer $7x$ is equal to one of the numbers $0$ or $1$, i.e., $x\in \{0,1/7\}$. Similarly, $y,z\in \{0,1/7\}$.

At this point we have only $2^3=8$ triples $(x,y,z)$, which are candidates to solve the problem, so we could test them individually. However, this testing can be avoided by noting that if any two of the numbers $x,y,z$ were equal to $1/7$, one of the expressions $3x+5y$, $3y+5z$, $3z+5x$ would be $8/7$, which is greater than $1$, and that is a contradiction. So, at most one of the numbers $x,y,z$ is equal to $1/7$ and the others are equal to zero. But then each of the three (non-negative) expressions $3x+5y$, $3y+5z$, $3z+5x$ is at most equal to $5/7$, so the conditions stated in the beginning of the solution as equivalence are satisfied and all such triples are solutions.

Conclusion. The problem has exactly $4$ solutions $$(x,y,z)\in \left\{(0,0,0),\left(\frac{1}{7},0,0\right),\left(0,\frac{1}{7},0\right),\left(0,0,\frac{1}{7}\right)\right\}.$$