# Prove that $m=[\sqrt{n}+(1/2)]$ for a given conditions

For any real number $x$, let $[x]$ denote the largest integer which is less than or equal to $x$.

Let $N_1=2$, $N_2=3$, $N_3=5$, and so on be the sequence of non-square positive integers. If the $n$th non-square positive integer satisfies $m²<N_n<(m+1)²$, then show that $m=[\sqrt{n}+(1/2)]$.

There are $(m+1)^2-m^2-1=2m$ integers $N_n$ strictly between $m^2$ and $(m+1)^2$. It follows that there are $$2+4+\ldots+2(m-1)=(m-1)m$$ such integers $<m^2$. This means that $m^2<N_n<(m+1)^2$ implies $$(m-1)m<n<m(m+1)\ .\tag{1}$$ Since all entries in $(1)$ are integers we then also have $$(m-1)m<n+{1\over4}<m(m+1)\ ,$$ or $$\left(m-{1\over2}\right)^2<n<\left(m+{1\over2}\right)^2\ .$$ From this we conclude that $$m<\sqrt{n}+{1\over2}<m+1\ ,$$ or $$m=\left\lfloor \sqrt{n}+{1\over2}\right\rfloor\ .$$