Find $m,k\in \mathbb N$ such that : $1000=m+(m+1)+(m+2)+...+(m+k)$ Problem :
Find $m,k$ be a natural number  such that : $1000=m+(m+1)+(m+2)+...+(m+k)$
My try : 
$m+(m+1)+(m+2)+...+(m+k)=2m(k+1)+k(k+1)$ 
$=(k+1)(2m+k)$ 
Now if $k$ odd then $k+1$ even also $2m+k$ odd 
If $k$ even then $k+1$ odd so $2m+1$ even 
so I will writte : 
$1000=125.4$ , $1000=4.125$ $....$
But how I find this two number ??
There many Case ??
 A: You've made a slight mistake.
You should have:
$m+(m+1)+(m+2)+...+(m+k)=m(k+1)+\frac 12 k(k+1)$ 
$=\frac 12(k+1)(2m+k)$
Set this to be equal to 1000:
$\frac 12(k+1)(2m+k)=1000$
You are right that there are many possible values of $m$ and $k$, so a good idea is to write one of them in terms of the other. It's easier to make $m$ the subject because it only appears once.
$\frac 12(k+1)(2m+k)=1000$
$(k+1)(2m+k)=2000$
$2m+k=\frac {2000}{k+1}$
$2m=\frac {2000}{k+1}-k$
$2m=\frac {2000}{k+1}-\frac{k(k+1)}{k+1}$
$2m=\frac {2000-k^2-k}{k+1}$
$m=\frac {2000-k^2-k}{2(k+1)}$
You are then free to choose any value of $k \in \mathbb Z$ as long as $k \ge 1$ and you use the formula to find the corresponding value of $m$.
A: As shown by @tomi, 
$$(k+1)(2m+k)=2000=2^45^3$$ so that there are a priori $(4+1)(3+1)=20$ decompositions to try.
But the two factors have opposite parities and the first is smaller so that only $16\cdot125$, $5\cdot400$ and $25\cdot80$ are possible. Hence $k=15,m=55$ or $k=4,m=198$ or  $k=24,m=28$.
