
Problem 1. Find all natural numbers $a, b$ and $c$ such that the roots of the equation

$$
\begin{array}{r}
x^{2}-2 a x+b=0 \\
x^{2}-2 b x+c=0 \\
x^{2}-2 c x+a=0
\end{array}
$$

are natural numbers.


Solution: Let $\left\{x_{1}, x_{2}\right\},\left\{x_{3}, x_{4}\right\}$ and $\left\{x_{5}, x_{6}\right\}$ be the roots of the first, second and third equation respectively, and let all of them be natural numbers.

Assume that $x_{i} \geq 2$ for all $i=1,2, \ldots, 6$. Then $2 a=x_{1}+x_{2} \leq$ $x_{1} x_{2}=b, 2 b=x_{3}+x_{4} \leq x_{3} x_{4}=c$ and $2 c=x_{5}+x_{6} \leq x_{5} x_{6}=a$. Thus $2(a+b+c) \leq a+b+c$, which is impossible, since $a, b, c$ are natural numbers.

Therefore at least one of the numbers $x_{i}$ equals 1 . Without loss of generality suppose $x_{1}=1$, so $1-2 a+b=0$.

If $x_{i} \geq 2$ for $i=3,4,5,6$, then

$$
2(b+c)=\left(x_{3}+x_{4}\right)+\left(x_{5}+x_{6}\right) \leq x_{3} x_{4}+x_{5} x_{6}=c+a
$$

whence $2(2 a-1+c) \leq c+a \Rightarrow c \leq 2-3 a$, which is impossible when $a, b, c$ are natural numbers.

So at least one of $x_{3}, x_{4}, x_{5}, x_{6}$ equals 1. Let $x_{3}=1$. Now $1-2 b+$ $c=0$. Assuming that $x_{5} \geq 2$ and $x_{6} \geq 2$, we get $2 c=x_{5}+x_{6} \leq a$, so $2(2 b-1) \leq \frac{b+1}{2}$. Thus $7 b \leq 5$, a contradiction.

Therefore at least one of the numbers $x_{5}, x_{6}$ is 1 and it follows that $1-2 c+a=0$. Further

$$
0=(1-2 a+b)+(1-2 b+c)+(1-2 c+a)=3-(a+b+c)
$$

and since $a, b, c$ are natural numbers, it follows that $a=b=c=1$.

Direct verification shows that $a=b=c=1$ satisfy the conditions of the problem.
