, 2015

Suppose that the fraction $\frac{3n-1}{2n^2+1}$ is reducible. Then there exists a positive integer $a$ different from $1$ which divides $3n-1$ and $2n^2+1$. It follows that $a$ divides also $3(2n^2+1)-2n(3n-1) = 2n+3$ and hence also $3(2n+3)-2(3n-1) = 11$. Since $11$ is a prime number it follows $a=11$, hence $11$ divides $3n-1$ and $2n^2+1$. Therefore there exists an integer $k$ such that $3n-1=11k$. From this we express $n = \frac{11k+1}{3}$. For this to be an integer, $3$ must divide $11k+1$ and hence $2k+1$, which can happen if and only if $k$ is of the form $k = 3m+1$ for some integer $m$. In this case we have $n = 11m+4$ and the number $2n^2+1 = 2(11m+4)^2+1 = 2 \cdot 11^2m^2 + 4 \cdot 11m + 33$ is also divisible by $11$. Because of $1 \le n \le 2015$ we have $0 \le m \le 182$. The given fraction is thus reducible for $183$ positive integers $n$.