4. Let three non-zero natural numbers $x, y$, and $z$, with $x>z$.

a) If $7x - 5y + 28z = 0$, then the number $y(x-z)$ is divisible by 35.

b) Determine $x, y$, and $z$ knowing that they are prime numbers less than 20, such that the number $y(x-z)$ is divisible by 35.

Dorina Bocu
## Solution.

a) From $7(x+4 z)=5 z$ it follows that $7 \mid(5 y)$. But $(5,7)=1$, so $7 \mid y$. (1p)

We have $5(y-7 z)=7(x-z)$, from which $5 \mid[7(x-z)]$. But $(5,7)=1$, so $5 \mid(x-z)$. (1p)

From the two relations, it follows that $y(x-z)$ :35. (1p)

b) Since, based on the hypothesis, $x-z<35$, it must be that $5 \mid y$ or $7 \mid y$. Since $y$ is prime, it follows that $y \in\{5,7\}$. (1p)

If $y=5$, then $7 \mid(x-z)$, from which $(x-z) \in\{7,14\}$. We obtain the solutions $(17,5,3)$ and $(19,5,5)$.

If $y=7$, then $5 \mid(x-z)$, from which $(x-z) \in\{5,10,15\}$. We obtain the solutions $(7,7,2),(17,7,7)$ and $(17,7,2)$. (3p)