3. If a, b, c are distinct positive real numbers, what is the smallest possible number of distinct numbers among the numbers $a+b, b+c, c+a, ab, bc, ca, abc$?

(Patrik Bak)
SOLUTION. Since $a, b, c$ are distinct positive numbers, so are the numbers $a b, b c, c a$, because, for example, from $a b=b c$ it follows that $a=c$ (thanks to $b \neq 0$). We see, then, that in the examined septet of numbers $a+b, b+c, c+a, a b, b c, c a, a b c$ there are at least 3 different values. We will first prove by contradiction that exactly 3 values can never be. Then we will provide an example of the examined septet, which is composed of only 4 different values.

First, let us assume that in some septet there are exactly 3 different values. We know that these are the values of the three products $a b, b c, c a$, and so the product $a b c$ must equal one of them. This means that one of the numbers $a, b, c$ is equal to 1, because, for example, from $a b c=a b$ it follows that $c=1$.

Without loss of generality, we will further limit ourselves to the case $c=1$. The septet in question with three different values can then be reduced to a sextet of the same property, which is composed of the numbers

$$
a+b, a+1, b+1, a b, a, b
$$

(we have substituted $c=1$ and omitted the number $a b c$ equal to $a b$). Since the equalities $a=1$ and $b=1$ have already been ruled out, the three different values are represented both in the first triplet $a+b, a+1, b+1$ and in the second triplet $a b, a, b$. Both numbers $a, b$ from the second triplet, therefore, must lie in the set $\{a+b, a+1, b+1\}$. This can only be achieved if $a=b+1$ and simultaneously $b=a+1$, which is impossible. The proof by contradiction is complete.

As we have promised, in the second part of the solution, we will provide an example of the examined septet, which is composed of 4 different values. According to the previous observations, it is worthwhile to examine the situation where, for example, $c=1$ and at the same time $b=a+1$. Then we have

$(a+b, b+c, c+a)=(2 a+1, a+2, a+1) \quad$ and $\quad(a b, b c, c a, a b c)=\left(a^{2}+a, a+1, a, a^{2}+a\right)$.

It is enough to find such a positive number $a \neq 1$ so that in the quintet of numbers

$$
a, a+1, a+2,2 a+1, a^{2}+a
$$

there are only four different values. Considering the obvious inequalities that hold between these five numbers for the considered $a$, the requirement can be met in exactly two ways. They are expressed by the equations

$$
a^{2}+a=2 a+1, \quad \text { resp. } \quad a^{2}+a=a+2 .
$$

Both actually lead to suitable triplets, which are of the form

$$
(a, b, c)=\left(\frac{1+\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}, 1\right), \quad \text { resp. } \quad(a, b, c)=(\sqrt{2}, 1+\sqrt{2}, 1) \text {. }
$$

Conclusion. The smallest possible number of different numbers in the examined septet is equal to 4.

Note. If we do not distinguish triplets $(a, b, c)$ that differ only in the order of their elements, there are two more admissible triplets for which only 4 different values can be found in the examined septet. The first of them is the triplet

$$
(a, b, c)=\left(a, \frac{a}{a-1}, \frac{a}{(a-1)^{2}}\right)
$$

where $a>0$ is the only real root of the cubic equation $a^{3}-4 a^{2}+4 a-2=0$. The second suitable triplet is

$$
(a, b, c)=\left(a, \frac{a}{a^{2}-a-1}, \frac{a(a-1)}{a^{2}-a-1}\right)
$$

where $a$ is the larger of the two positive roots of the equation $a^{4}-2 a^{3}-2 a^{2}+2 a+2=0$.