How many...

Let $Q = \dfrac {8n+46}{2n+1} = \dfrac {4(2n+1)+42}{2n+1} = 4 + \dfrac {42}{2n+1}$ . For $Q$ to be a positive integer $\dfrac {42}{2n+1} = \dfrac {2\times 3 \times 7}{\color{#3D99F6}2n+1}$ must be a positive integer.

Since $\color{#3D99F6}2n+1$ is odd, we have

${\color{#3D99F6}2n + 1}= \begin{cases} \color{#3D99F6}3 & \implies \dfrac {2 \times \cancel 3 \times 7}{\color{#3D99F6}\cancel 3} = 14 & \implies n = 1 \\ \color{#3D99F6} 7 & \implies \dfrac {2 \times 3\times \cancel 7}{\color{#3D99F6}\cancel 7} = 6 & \implies n = 3 \\ \color{#3D99F6} 21 & \implies \dfrac {2 \times \cancel 3 \times \cancel 7}{\color{#3D99F6}\cancel {21}} = 2 & \implies n = 10 \end{cases}$ .

Therefore, there are $\boxed 3$ positive integers $n$ satisfy the condition.

Suppose n is an integer for which $2n + 1$ is a divisor of $8n + 46$ , then there is an integer $k$ such that: $\begin{aligned} k(2n + 1)&=&8n + 46\\ k(2n + 1)&=&4(2n+1)+42\\ (k-4)(2n + 1)&=&2\cdot 3\cdot 7 \end{aligned}$

As $n$ is restricted to be positive the only possibilities for $2n+1$ are to be 3 or 7 or 21 that give $\boxed{\mbox{three}}$ posibilities for $n$ namely 1 or 3 or 10.