Relate, a perfect square $m^2$ with the $n-th$ non-square positive integer.

Question

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)] $.

I searched, this site, and found this question answered. It was answered by the user, Christian Blatter. But I never, understood the solution. I am a new user, and I don't yet have the privilege to comment everywhere as of yet. (So, I thought about asking for clarification, here in the hope that this doesn't violate the site policy). I hereby, post his answer :

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 $$m=\left\lfloor \sqrt{n}+{1\over2}\right\rfloor\ .$$

Now, I don't understand, how did he claim, " $m^2<N_n<(m+1)^2$ implies $$(m-1)m<n<m(m+1)\ .\tag{1}$$ "? Moreover, I feel

From this we conclude that $$m<\sqrt{n}+{1\over2}<m+1\ ,$$ or $$m=\left\lfloor \sqrt{n}+{1\over2}\right\rfloor\ .$$

is an incorrect implication as this is not in agreement with the definition of $[x],$ as:

$[x]=n$ iff, $n\leq x \leq n+1$ (where $n$ is an integer.)

So, we can't really say, $$m=\left\lfloor \sqrt{n}+{1\over2}\right\rfloor\ ,$$ unless $$m\leq\sqrt{n}+{1\over2}<m+1\ ,$$ is true, instead of $$m<\sqrt{n}+{1\over2}<m+1\ .$$ Am I missing something important ?

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

Answer 1

For your first question, $(1)$, the idea is he is counting the amount of non-square numbers below $m^2$, and gets that there are $(m-1)m$ such numbers. Since $N_n$ is greater than $m^2$, we see that the index $n$ must be greater than $(m-1)m$, since $N_i$ for $i\leq (m-1)m$ must satisfy $N_i<m^2$. The upper bound on $n$ is completely similar (I missed a detail, see update below).

For your second question, remember that a strict inequality is the same as a weak inequality with extra information. So $a<b$ is the same as $a\leq b$ and $a\neq b$. Thus, you can definitely conclude from the strict inequalities the desired result.

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The upper bound in the proof should not be strict, i.e. we have $n\leq (m+1)m$ but are not guaranteed $n<(m+1)m$ immediately. This is because there are $(m+1)m$ non-square numbers $N_i$ satisfying $N_i<(m+1)^2$, and $N_n$ could be the final one, as you point out in your comment. But the weak inequality is enough to finish the proof, though we have to be a bit more careful.

The two following lines in the proof still hold, but with the right inequality being weak, and so we arrive at $m<\sqrt{n}+\frac12\leq m+1$. Here we cannot a priori conclude our desired result. But, recalling what we know about square roots, we can actually achieve a strict inequality anyway.

First, if $n$ is a perfect square, say $k^2 = n$, we get $m<k+\frac12\leq m+1$. Since $m+1$ is an integer and $k+\frac12$ is not, they cannot be equal, and so we get our strict inequality.

If $n$ is not a perfect square, then $\sqrt{n}$ is irrational, and so the same is true for $\sqrt{n}+\frac12$. Thus, it cannot be an integer, in particular not $m+1$, and so we get strict inequality again. This fixes the proof.