Least value of $C$ for which $CT_n+1$ is always a triangular number. 
For $n \geq 1$, the $n$th triangular number is defined as $T_n = 1+2+\cdots+n$. Let $C$ be the least positive integer such that $CT_n + 1$ is a triangular number for all positive integers $n$. How many factors does $C$ have?

Attempt
We have that $T_n = \dfrac{n(n+1)}{2}$ and $CT_n+1 = \dfrac{Cn(n+1)}{2}+1 = \dfrac{Cn(n+1)+2}{2} = \dfrac{k(k+1)}{2} \implies Cn(n+1)+2 = k(k+1)$. We must find a $C$ such that for every $n$ there exists an integer $k$ that satisfies the relation. How do I proceed?

Continuing your attempt: If $Cn(n+1)+2 = k(k+1) = (k+\frac12)^2 - \frac14$, then $Cn^2+Cn+2+\frac14 = (k+\frac12)^2$ is a square. This implies that the quadratic equation (in $n$)
$$Cn^2 + Cn + \frac94 = 0\tag1$$
has two identical roots, so the discriminant of (1) is zero:
$$
C^2 - 9C = 0\tag2
$$
The only positive solution to (2) is $C=9$.
As a check, with $C=9$ we have  $Cn(n+1)+2=(3n+1)(3n+2)$.
