What properties should a number $n$ have so there is an integer $k$ where $n = k \cdot (k+1)$? This is my first question here so please bear with me:)
What properties should a natural number $n$ have so there is an integer $k$ where $n = k \cdot (k+1)$? By properties, I mean things such as divisibility, prime factors, etc. I tried using quadratic equations but couldn't come up with anything.
Thank in advance!
 A: I think this is the right answer:
If $n = k \cdot (k+1)$, then $-k^2 - k + n = 0$ and using the quadratic equation, we have $k = \frac {1 \pm \sqrt{1 + 4n}}{-2}$ so $1 + 4n$ needs to be a perfect square.
A: You can view it as a quadratic equation, which can be solved
$$n=k(k+1)\\k^2+k-n=0\\
k=\frac 12\left(-1\pm\sqrt{1+4n} \right)$$
So you need $1+4n$ to be a square.  The square root will be odd, so the division by $2$ is no problem.  The two signs give one solution with $k \gt 0$ and the other with $k \lt 0$, negating both $k,k+1$
A: An integer $n$ has this property iff
$$n=\lfloor \sqrt n\rfloor(\lfloor \sqrt n\rfloor +1) $$
A: Well, the condition that it is equal to some $k(k+1)$ IS a condition that will make this true.
You can list these numbers by letting $k$ be any integer and letting $n = k(k+1)$
You can determine if an arbitrary number $n$ has this condition by finding $k$ so that $k^2 \le n < (k+1)^2$ (there will always be a unique such integer for all natural $n$; we could write this as $k=\lfloor \sqrt n \rfloor$) and checking whether $n-k^2 = k$ or not.
If $n - (\lfloor \sqrt n \rfloor)^2 = \lfloor \sqrt n \rfloor$ then $n$ has the condition and if not then $n$ doesn't.
But that's not really an answer as that's just another way of saying $n = k^2 +k = k(k+1)$.  i.e. it is true when it is true.  But it does tell you how to check if it is true.
