$[\sqrt{m}]=[\sqrt{m+125}]$--> $\sqrt{m+125}-\sqrt{m}<1$---> $m>62^2$ ---> $m=3845,...,5000$ answer 1156

anybody??

We have for some integer $k$, $$k\le\sqrt m\text{ and }\sqrt{m+125}<k+1\implies k^2\le m<k^2+2k-124\implies 63\le k\le70$$The answer is $$\sum_{k=63}^{70}(2k-124)=\boxed{72}$$

That's correct!!