Does the absence of horizontal lines shows that there are no $n,m\in \mathbb{N}$ such that $n^2=2m^2$? When I was learning about the proof of the irracionality of $\sqrt{2}$, I remember of trying to visualize it by ploting the graphs of $f(n)=n^2$ and $g(m)=2m^2$, but at the time I got confused and obtained nothing out of these.
The problem is about finding integers $n,m\in \mathbb{Q}$ such that $n^2=2m^2$, then I assumed that they can't be found and then I thought: What would happen if the could be found? And then I created one example in which the integers can be found, I chose $2n=4m$ and made a plot with $h(n)=2n$ and $p(m)=4m$, I've obtained this:



I've noted that whenever I can draw a horizontal lime, there is a pair of integers such that $2n=4m$, then I tried to make the same process with $f(n)=n^2$ and $g(m)=2m^2$ and I obtained this:



And I wanna be sure about one thing: Does the impossibility of finding horizontal lines perpendicular to the $x$-axis connecting the integral points in the functions mean that it's impossible to find $n,m$ (excluding the trivial $n=m=0)$ integers such that $n^2=2m^2$? 
 A: You consider a horizontal line such that $n^2 = 2m^2$.
Note that $n^2$ is even and therefore $n$ is even.
Suppose we start and move the horizontal line upwards. If we find a line - we are done.
We are now at a position that we have not found a line yet.
Suppose we have found a horizontal line that matches $n^2 = 2m^2$.
As $n$ is even, $m$ cannot be even, otherwise we would have found a horizontal line already for half the values, as $(2n)^2 = 2(2m)^2 \Rightarrow n^2 = 2 m^2$.
So $n$ is even and $m$ is odd. Thus we can write $n=2k$ and $m=2\ell+1$,
whence we obtain
$$
2 k^2 = 4\ell^2+2\ell+1,
$$
but the left side is even and the right side is odd, so we cannot find such number, i.e. we cannot find such a horizontal line by moving the line upwards.

Another proof:
Suppose we have a horizontal line such that $n^2 = 2m^2$.
Then due to the grid we can write $n=m+a$, where $a$ is an integer, and we obtain
$$
(m+a)^2 = 2m^2 \Rightarrow (m-a)^2 = 2a^2,
$$
however $m-a<m+a$, so there is a lower horizontal line.
As for every horizontal line there exists a lower horizontal line, there is no lowest horizontal line, unless $m+a=m-a$, thus $a=0$, i.e. $m=0$ and $n=0$ which is the trivial case. Consequently, the only horizontal line is the trivial case...
A: Yes, your reasoning is correct. However I don't see how this can be turned into a proof... I don't see a geometric evidence of the fact the no two points are at the same height.
