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.