B1. Find all integers $z$ for which $\frac{5 z^{2}+3}{z-1}$ is also an integer.

(20 points)
B1. The expression $\frac{5 z^{2}+3}{z-1}$ is first transformed as follows:

$$
\frac{5 z^{2}+3}{z-1}=5 z+\frac{5 z+3}{z-1}=5 z+5+\frac{8}{z-1}
$$

Since $z$ is an integer, $\frac{8}{z-1}$ must also be an integer. This means that $z-1$ must be one of the divisors of the number 8. The divisors of 8 are $-8,-4,-2,-1,1,2,4$ and 8, so $z \in\{-7,-3,-1,0,2,3,5,9\}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_cfd2cf88ee04390f062cg-12.jpg?height=68&width=1579&top_left_y=1071&top_left_x=273)

![](https://cdn.mathpix.com/cropped/2024_06_07_cfd2cf88ee04390f062cg-12.jpg?height=65&width=1576&top_left_y=1121&top_left_x=274)
The realization that $\frac{8}{z-1}$ must be an integer .........................................................

![](https://cdn.mathpix.com/cropped/2024_06_07_cfd2cf88ee04390f062cg-12.jpg?height=55&width=1576&top_left_y=1229&top_left_x=274)

![](https://cdn.mathpix.com/cropped/2024_06_07_cfd2cf88ee04390f062cg-12.jpg?height=60&width=1579&top_left_y=1278&top_left_x=273)

![](https://cdn.mathpix.com/cropped/2024_06_07_cfd2cf88ee04390f062cg-12.jpg?height=60&width=1579&top_left_y=1329&top_left_x=273)