Can someone find a solution to this problem:
You try to build an equilateral triangle with a number of pearls. Your pearls are placed the following way in order to solve the triangle:
$1$ pearl in the first row, $2$ in the second row, $3$ in the next, and so on, with as many rows as you may need to form an equilateral triangle.
In the first round you get $2008$ pearls to try and form this equilateral triangle and you see that it doesn‘t work. In the next round you get one more pearl, so $2009$, and again it does not work. Then you get two pearls, so $2011$, and it still doesn‘t work. Every new round you get the amount of stones you got in the round before and $1$ more than that. (so after the try with $2011$ we get the $2+1=3$ pearls added to $2011$ for the next try).
How many equilateral triangles can you build this way, if you continue to get more and more pearls until the end of time.
So far I know that the amount of pearls we gain in addition to $2008$ is an arithmetic series with the difference of $1$ between its consecutive terms and so after $n$ rounds we have $2008 +\dfrac{n(n+1)}{2}$ pearls.
Where do I go from here?
We can form a complete equilateral triangle only when the total number of pearls is a Triangular number (that is, expressible as $1+2+3+\ldots$).
So we're looking for solutions of $$2008 + \frac{n(n+1)}{2}=\frac{m(m+1)}{2}$$ for positive integers $m,n$. On rearranging, $$m^2-n^2+m-n=2008\cdot 2$$ $$\Rightarrow (m-n)(m+n+1)=4016=2^4 \cdot 251$$
which is easily solvable by observing that $m-n$ and $m+n$ have same parity for all integers $m,n$.
One solution would be $m=133$, $n=\ldots$
And there would be others.