b/w....

Total numbers lying between $n^{2}$ and $[n+1]^2$ is 2 times n. Therefore $99999 \times 2 = 199998$ is the answer.

Excluding the numbers, there are (1000000^2 - 99999^2) - 1 = [(1000000+99999) (1000000-99999)] - 1 = 199999 1 - 1 = 199998

no. in between can be find by

$(100000^{2} - 99999^{2})-1$

$(100000-99999)(100000+99999)-1$

$(1\times199999) -1$

hence 199998...1 is subtracted in order to exclude one of the given no.

the ans will be 100000^2 -99999^2 -1 =199999-1=199998