I am trying to prove that there exists a perfect square between a natural number and its double, and I am trying to prove it by induction. Here is my proof so far, but I am kind of stuck.
for $n=1$, choose $m=1$. For $n=2$, choose $m=2$.
I figured out that when $ m^2 \geq n+1$ then the the square overlaps between the range from $n$ to $2n$ and $n+1$ to $2(n+1)$.
So I need to prove it when $m^2=n$.
So $n\leq m^2$
$n+1 \leq m^2+1 \leq m^2 + 2m + 1 = (m+1)^2$
but now I can't prove that
$(m+1)^2 \leq 2n+2$
The furthest I got is trying to prove that
$ m^2 + 2m \leq 2n$
Note that when $m\ge 2$ and $m^2 = n$,
$$(m+1)^2 = m^2 + 2m + 1 \le 2m^2 +1 \le 2m^2 + 2 = 2n+2.$$