An olympiad question

Question

I came across a question ( mentioned below ) which is from a Mathematical Olympiad.

It says,

If $m $ and $n $ are positive integers such that

$n + (n + 1) + (n + 2) +...+ (n + m) = 1000 $

then how many $(m, n)$ pairs exists ?

Answer 1

$n(m+1)+(1+2+\cdots+m)=1000$

$n(m+1)+\dfrac 12m(m+1)=1000$

$(m+1)(2n+m)=2000$

Note that $m+1$ and $m+2n$ are of opposite parities. We have to factorize $2000$ into the product of an odd and an even number. Note that both factors should be greater than $1$ as $m$ is positive.

$2000=5\times400=25\times80=125\times16$

Note also that $m+1<2n+m$.

So, $(m+1,2n+m)=(5,400)$ or $(25,80)$ or $(16,125)$

$(m,n)=(4,198)$ or $(24, 28)$ or $(15,55)$.