Square numbers between n and 2n -- Check my proof please

[Raffaele]

I noticed that for any integer greater than 4 there exists at least one perfect square in the open interval (n, 2n). I think I have proved the statement, but as I am not a professional, I'd like someone to review my proof.

By induction we see that it is true for $n=5$ because $5<9<10$.

Now suppose that it is true for n, and let's prove it for n + 1.
From the inductive hypothesis we know that there is an x such that $n<x^2<n+1$. This square should work for n + 1 too. Like in the case of $n=5$ the square 9 works for 6, 7 and 8 too. But not for $n=9$ or greater. Here we need the next square 16.

So if $x^2\ge n+1$ we must show that $(x+1)^2$ is the square we need. In other words we have to verify that
$$n+1<(x+1)^2<2(n+1)$$
Notice that for positive n, if $x^2\le n+1$ then $x\le \sqrt{n+1}$. As $n<x^2$ it follows that
$$n+1<x^2+1<x^2+2x+1=(x+1)^2$$
We have to show that $(x+1)^2<2(n+1)$
$$(x+1)^2=x^2+2x+1<\left(n+1\right)+2\sqrt{n+1}+1$$
$$\left(n+1\right)+2\sqrt{n+1}+1 < 2(n+1)$$
$$n^2-4n-4>0\rightarrow n>2\sqrt 2 + 2\rightarrow n\ge 5$$
So we proved that for any $n\ge 5$ there is at least one x such that $n< x^2 < 2n$.

 

[mathman]

The statement is true, but the proof looks more complicated than necessary - you don't need mathematical induction. All you need is (similar to what you have) is $n\gt 2\sqrt{n} + 1$ for n > 4.

 

[Simon Bridge]

Looks like Your 4th sentence is false to start with.
$x^2 \geq n+1$ ... consider n=8, interval is 9,10,11,12,13,14,15,16,17 ... there are two squares in the interval.
7th sentence... try verifying the statement by example... n=5
Then the expression asserts that 6<100<12 ... which is false.

At best you need to tidy up your notation.

By induction, you want to assume: $\exists x\in\mathbb Z: n <x^2<2n$ is true, then prove
$\exists x\in\mathbb Z: n+1 <x^2<2n+2$ .

 

[pbuk]

No, you want to assume: $\exists x \in\mathbb Z: n <x^2<2n$ is true, then prove $\exists y \in\mathbb Z: n+1 <y^2<2n+2$ . Most of the time you can take y=x and this is trivial, but it doesn't always work.

 

[PeroK]

In fact, you can always find $x$ such that:

$\sqrt{n} < x \le \sqrt{n}+1$

This $x$ does the job unless $6n \ge n^2 + 1$, which only holds for $n < 6$

Then you can check $n = 5$ manually.

 

[mathman]

My previous note is in error. The key expression should be $\lfloor \sqrt{n} \rfloor +1 \lt \lceil \sqrt{2n} \rceil$.