1. For how many natural numbers $n, n \leq 2015$, is the fraction $\frac{3 n-1}{2 n^{2}+1}$ not in its simplest form?
III/1. Denote the fraction $\frac{3 n-1}{2 n^{2}+1}$ is not simplified. Then there exists a natural number $a$, different from 1, that divides both $3 n-1$ and $2 n^{2}+1$. It follows that $a$ also divides $3\left(2 n^{2}+1\right)-2 n(3 n-1)=2 n+3$ and therefore also $3(2 n+3)-2(3 n-1)=11$. Since 11 is a prime number, it follows that $a=11$, so 11 divides both $3 n-1$ and $2 n^{2}+1$. This means there exists an integer $k$ such that $3 n-1=11 k$. From this, we express $n=\frac{11 k+1}{3}$. For this to be an integer, 3 must divide $11 k+1$ or $2 k+1$, which happens exactly when $k$ is of the form $k=3 m+1$ for some integer $m$. In this case, $n=11 m+4$ and the number $2 n^{2}+1=2(11 m+4)^{2}+1=2 \cdot 11^{2} m^{2}+4 \cdot 11 m+33$ is also divisible by 11. Since $1 \leq n \leq 2015$, it follows that $0 \leq m \leq 182$. The given fraction is therefore not simplified for 183 natural numbers $n$.

Example of a fraction that is not simplified (e.g., $n=4$)

1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-15.jpg?height=51&width=1627&top_left_y=694&top_left_x=223)

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-15.jpg?height=52&width=1633&top_left_y=745&top_left_x=220)

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-15.jpg?height=49&width=1631&top_left_y=798&top_left_x=218)

![](https://cdn.mathpix.com/cropped/2024_06_07_77e56b77a012e45bd48fg-15.jpg?height=52&width=1631&top_left_y=842&top_left_x=218)

Conclusion that the fraction is not simplified for 183 numbers ..............................................................