1. $a, b, c, d$ are distinct positive integers for which

$$
a+b=c d
$$

and the

$$
a b=c+d
$$

equality holds. Determine all such quadruples of numbers.
I. solution. Since the role of the four numbers is not distinguished from each other, we can assume without loss of generality that $a$ is the smallest, and furthermore, that $ca+b=c d \geq 3 d>c+2 d=a b+d>a b$
$$
\text{must hold, but this is impossible. Therefore}
$$
$$
a=1, \quad c \geq 2 \quad \text { and } \quad d \geq c+1
$$
thus
$$
a+b=1+b=c d \geq 2 d \geq c+1+d=a b+1=b+1
$$
This can only hold if the equality sign is valid everywhere, so
$$
c=2, \quad d=c+1=3, \quad \text { and } \quad b=c d-1=5
$$
The number pairs 1, 5 and 2, 3 indeed meet the conditions. From these, we can obtain 7 more suitable number pairs if we swap the elements of the pairs with each other, and if we swap the two pairs.

II. solution. Transform the product $(a-1)(b-1)$ using the conditional equalities:

$$
(a-1)(b-1)=a b-a-b+1=c+d-c d+1=2-(c-1)(d-1)
$$
that is
$$
(a-1)(b-1)+(c-1)(d-1)=2
$$
Since the problem is about positive integers, this can only hold if both terms on the left are 1, or one is 0 and the other is 2. The first possibility only follows in the case of $a=b=c=d=2$. These satisfy the equalities required by the problem, but the numbers are not distinct.

The second case follows when one of the numbers is 1, say $a=1$, from which it follows that
$$
(c-1)(d-1)=2
$$
We can assume that $c1
$$
it cannot be $a>1$, because then $b>2$, and thus
$$
\frac{1}{a}+\frac{1}{b} \leq \frac{1}{2}+\frac{1}{3}2$, because then $d>3$, and thus

$$
\frac{1}{c}+\frac{1}{d} \leq \frac{1}{3}+\frac{1}{4}<\frac{2}{3}
$$
Therefore $c=2$ and the conditional equations are thus formulated:

$$
b=1 \cdot b=2+d, \quad 1+b=2 d=2 b-4
$$
so $b=5, d=3$ The number pairs 1,5; 2,3 meet the conditions.

Remark. After we have established that $a=1$, an interesting conclusion is the following: from the equalities

$$
1+b=c d, \quad b=c+d
$$
multiplying the second by $c$ and using the first
$$
b c=c^{2}+1+b
$$
Multiplying the equality by 4, we transform it as follows:

$$
4 c^{2}-4 b c+b^{2}+8=b^{2}-4 b+4, \quad \text { that is } \quad(2 c-b)^{2}+8=(b-2)^{2}
$$
The difference between the square numbers is 8 only between 1 and 9, so $|2 c-b|=1,|b-2|=3$, and from this we again obtain the two number pairs mentioned earlier.

IV. solution. From the conditional equations, it follows that

$$
\frac{a b}{a+b}=\frac{c+d}{c d}
$$
Both fractions are positive, so either $\frac{a b}{a+b}$ and $\frac{c+d}{c d}$ are both 1, or the first, say, is less than 1, and the second is greater than it. In the first case, $a b=a+b$, from which $(a-1)(b-1)=1$, and thus - since we are talking about positive integers $a=b=2$. However, these are not distinct.

In the second case, $a b<a+b$, and since we are talking about integers,
$$
a b \leq a+b-1, \quad \text { that is } \quad(a-1)(b-1) \leq 0
$$
The left side cannot be negative, so its value is 0, that is, one of the numbers, say $a=1$.

Then
$$
c d=1+b=1+1 \cdot b=1+c+d
$$
expressing $c$ from this
$$
c=\frac{d+1}{d-1}=1+\frac{2}{d-1}
$$
Since $c$ is an integer, $d$ can only be 2 or 3, and the corresponding values of $c$ are 3 and 2, respectively. This again leads us to the number pairs 1, 5 and 2, 3.