<s>[INST] 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?
 [/INST] If
$m = \frac{n(n+1)}{2}
$,
$2m = n(n+1)
=n^2+n
$
so
$8m = 4n^2+4n
$
or
$8m+1 = 4n^2+4n+1
=(2n+1)^2
$.
For the reverse way,
if $8m+1$
is a square,
it must be the square
of an odd number,
so
$8m+1 = (2n+1)^2$,
and just do the above in reverse.
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