4A. Given the statements:

a) The number $x+1$ is divisible by $y$,

b) $x=2y+5$,

c) The number $x+y$ is divisible by 3,

d) $x+7y$ is a prime number.

Determine all pairs of natural numbers $(x, y)$ for which exactly three of the given statements are true, and one is false.
Solution: For any natural numbers $x$ and $y$, the number $x+7y$ can be represented as $x+7y = (x+y) + 6y$. According to this, the statements v) and g) cannot both be true at the same time. Therefore, the true statements are a) and b), and one of the statements v) or g). From the truth of b), it follows that for natural numbers $x$ and $y$, the equation $x+y = 3y + 5$ holds. Since 3 is not a divisor of $x+y$, statement v) is false. This means we need to determine all natural numbers for which statements a), b), and g) are true. From a) we have $x+1 = ky, k \in \mathbb{N}$, and from b) $x = 2y + 5$, i.e., $ky = 2y + 6$. We get $y = \frac{6}{k-2}, y \in \mathbb{N} \Rightarrow k-2 \in \{1,2,3,6\} \Rightarrow y \in \{1,2,3,6\}$, so we obtain the pairs: $(x, y) \in \{(17,6),(11,3),(9,2),(7,1)\}$, and from them, statement g) is satisfied only by the pairs (17,6) and $(9,2)$.

## 2nd Year

1 AB. Let $a, b$ and $c \neq 0$ be distinct real numbers. If the equations $x^{2} + ax + bc = 0$ and $x^{2} + bx + ca = 0$ have exactly one common root, then the other roots of these equations are roots of the equation $x^{2} + cx + ab = 0$. Prove!