Shortlisted problems for the 65th Romanian NMO

Let $a$, $b$, $c$ be arbitrary elements of $M$. By hypothesis, $ab + bc + ca$ is rational for all $a$, $b$, $c$ in $M$.

Let us fix $a$ and $b$ in $M$, and let $c$ vary over $M$.

For $c = a$, we have:

$$ab + bc + ca = ab + ba + aa = 2ab + a^2$$
which is rational.

For $c = b$, we have:

$$ab + bc + ca = ab + bb + ba = ab + b^2 + ba = 2ab + b^2$$
which is rational.

Now, consider the difference:

$$(2ab + a^2) - (2ab + b^2) = a^2 - b^2$$
which is rational. Therefore, $a^2 - b^2$ is rational for all $a$, $b$ in $M$.

Now, for $c$ arbitrary in $M$, $ab + bc + ca$ is rational. Fix $a$ and $b$, and let $c$ vary.

Let $c$ be arbitrary in $M$. Then $ab + bc + ca$ is rational.

Let us consider $ab + bc + ca$ as a function of $c$:

$$ab + bc + ca = ab + bc + ca = ab + c(b + a)$$
So, for fixed $a$, $b$, as $c$ varies over $M$, $ab + c(a + b)$ is rational.

Let $d = a + b$. Then, for all $c$ in $M$, $ab + cd$ is rational.

Let $c_1$, $c_2$ be two elements of $M$. Then $ab + c_1 d$ and $ab + c_2 d$ are both rational.

Their difference is $d(c_1 - c_2)$, which is rational. Since $d = a + b$ is a positive real, and $c_1$, $c_2$ are arbitrary in $M$, $c_1 - c_2$ is rational up to scaling by $d$.

But we already have that $a^2 - b^2$ is rational for all $a$, $b$ in $M$.

Let us now fix $a$, $b$ in $M$ and consider $a^2 - b^2$ is rational, so $a^2 = b^2 + r$ for some rational $r$.

Now, $a$, $b$ are positive reals, so $a = \sqrt{b^2 + r}$.

Let us now consider $\frac{a}{b}$.

Let $x = \frac{a}{b}$, then $a^2 = b^2 x^2$, so $a^2 - b^2 = b^2(x^2 - 1)$ is rational.

But $b^2$ is positive real, and $x^2 - 1$ is real. So $b^2(x^2 - 1)$ is rational for all $a$, $b$ in $M$.

Now, fix $b$ in $M$. Then for all $a$ in $M$, $b^2(x^2 - 1)$ is rational, where $x = \frac{a}{b}$.

But $b^2$ is a fixed positive real number, so $x^2 - 1$ is rational up to scaling by $b^2$.

Let $s = b^2(x^2 - 1)$ rational, so $x^2 = 1 + \frac{s}{b^2}$, so $x^2$ is rational.

Therefore, for all $a$, $b$ in $M$, $\left(\frac{a}{b}\right)^2$ is rational.

Now, $a$, $b$ are positive reals, so $\frac{a}{b}$ is positive real, and its square is rational. Therefore, $\frac{a}{b}$ is a positive real number whose square is rational, i.e., $\frac{a}{b}$ is a positive rational or a positive irrational whose square is rational.

But suppose $\frac{a}{b}$ is irrational and its square is rational, i.e., $\frac{a}{b} = \sqrt{q}$ for some positive rational $q$.

Let us check if this is possible. Suppose $M$ contains $a = b \sqrt{q}$ for some $b$ in $M$ and $q$ positive rational.

But then, for $a$, $b$ in $M$, $ab + bc + ca$ must be rational for all $c$ in $M$.

Let us check for $a = b \sqrt{q}$, $b$ in $M$, $c$ arbitrary in $M$.

$ab + bc + ca = b a + b c + c a = b a + c(b + a) = b a + c(b + a)$.

But $b a = b^2 \sqrt{q}$, $b + a = b + b \sqrt{q} = b(1 + \sqrt{q})$.

So $ab + bc + ca = b^2 \sqrt{q} + c b (1 + \sqrt{q}) = b^2 \sqrt{q} + c b + c b \sqrt{q} = c b + (b^2 + c b) \sqrt{q}$.

For this to be rational for all $c$ in $M$, $b^2 + c b$ must be zero unless $\sqrt{q}$ is rational, i.e., $q$ is a perfect square.

Therefore, $\frac{a}{b}$ must be rational for all $a$, $b$ in $M$.

Thus, $\frac{a}{b}$ is rational for all $a$, $b$ in $M$.