## Task 2 - 070922

For two rational numbers $a$ and $b$, the following four inequalities hold:

$$
a+b \neq 3 ; \quad a-b \neq 10 ; \quad a \cdot b \neq 5 ; \quad a: b \neq 18.75
$$

However, the numbers $3$, $10$, $5$, and $18.75$ match (in a different order) with each of the numbers $a+b$, $a-b$, $a \cdot b$, and $a: b$.

Determine the numbers $a$ and $b$!
Given that with $a=\frac{(a+b)+(a-b)}{2}>0$ also $b=\frac{a \cdot b}{a}$ is positive, it follows that $a-b<a+b$.

In particular, $a-b$ cannot be the largest of the four values. This leaves two cases: $a-b=5$ or $a-b=3$.

In both cases, $a-b \in \mathbb{N}$. If $a+b=18.75$, then $a \cdot b=\frac{(a+b)^{2}-(a-b)^{2}}{4} \notin \mathbb{N}$ would follow, because in the numerator of this fraction, a non-integer is subtracted from an integer, so the numerator itself is not an integer. But $a \cdot b \notin \mathbb{N}$ would be a contradiction, since the other three results are all natural numbers. Therefore, in both cases, $a+b \neq 18.75$.

1. Case: $a-b=5$.

Then the only remaining value for $a+b$ is 10, which leads to $a=\frac{15}{2}, b=\frac{5}{2}, a \cdot b=\frac{75}{4}=18.75$ and $a: b=3$, thus providing a solution.

2. Case: $a-b=3$.

If $a+b=10$, then $a=\frac{13}{2}, b=\frac{7}{2}$ and $a \cdot b=\frac{91}{4}$, which does not appear in the list. Therefore, the only remaining possibility to check is $a+b=5$, which leads to $a=4, b=1$ and $a \cdot b=4$, which is also not a solution.

Thus, there is exactly one solution pair, namely $(a, b)=\left(\frac{15}{2}, \frac{5}{2}\right)$.