If m is an integer and satisfy given two condition , then how many values of m are possible satisfying both condition.

$(1.) 1 \leq m \leq 5000$

$(2.) [ \sqrt{m} ] = [ \sqrt{m + 125} ] $

where [ x ] is the greatest integer function . Find how many values of m satisfy both these condition.

My attempt: By trial and error method I concluded that for any value of m less than $(63)^2$ doesnot satisfy given condition. Hence m=$3936,3970,3971$ where $63^2=3969$.Then again for m=$4906,4097,4098,4099$ so I'm not getting any proper trend. Is there any other way to solve it ?


Suppose $m = n^2$ where $n$ is an integer. In order to satisfy the second condition it's necessary to have $n^2+125<(n+1)^2 \Rightarrow 62\lt n$

That is, for $n\le62$ second condition won't hold so we just have to bother with $63^2=3969\le m\le 5000\le 5041=71^2$

If we replace $n^2$ by $n^2+k$ where $k$ is an integer greater than or equal to $0$ in the above we get $n^2+k+125<(n+1)^2 \Rightarrow 62+k/2 \lt n$

As we know that $n\ge 63$ let's replace $n$ by $63 + i$ where $i$ is defined as $k$

We get $62+k/2 \lt 63 + i \Rightarrow k\lt 2+2i$

Varying $i$ from $0$ to $7$ we see that $k$ can assume respectively $2,4,6,...,16$ values. Therefore the number of m's satisfying both conditions is $2+4+...+16=2(1+2+...+8)=2\frac{8(1+8)}{2}=72$


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