Prove that there exists $N$ different points $P_i$ such that not all of them are on the same line and their distance is $d(P_i,P_j) \in \Bbb N$ 
Let $N>2$ be a number ($N \in \Bbb N$) prove that there exists $N$ different points $P_i$ such that not all of them are on the same line and the distance between them is $d(P_i,P_j) \in \Bbb N$ for all $1 \leq i \leq j \leq N$


I always post my way organized and everything I tried but here I could not do it well as I have ideas in my mind but I do not know how to actually start this proof(the proof is not related to any topic I study it was just a fun question my professor gave).
first I tried to understand why it cannot be $N=1$ and thought that it is probably because of pythagorean theorem since we will get $1^2+1^2=C^2= \sqrt{2} \not \in \Bbb N$ and the same thing for $N=2$ but I am not sure that this is the reason as there are infinite numbers that do the same (where the result is not natural).
I thought about placing all the points on the $X$ axis except for oneto put it on point $(0,y)$ and from this point to pull a straight line to every other point and then maybe a triangle are will be an answer? or anything related to a triangle that I cannot quiet figure.
sorry for the picture but this is what I thought about:

I could not continue or even start the actual proof but this is my idea.
Note - I do not study this topic so if someone can help hopefully the explanation can be simple , Thank you for any tips and help
Edit - according to the comments the reason for $N=1,2$ is not possible because then they can be on the same line.
the question is related to pythogrean triples and the idea in the picture is correct.
$(a,b,c)=(2mn,m^2-n^2,m^2+n^2)$ will give us triples and the area of the triangle is $mn(m^2-n^2)$ but I am not able to put it in a generalized way and prove the question.. the idea is now understandable and more clear thanks to Jason DeVito but as mentioned I am not able to put it in a general way
 A: Claim. There is an infinite set of points on the unit circle such that the distance between any two of them is rational.
Proof. Parametrise by $(\cos\theta,\sin\theta)$ and consider the set $S$ of all points where $\tan\frac{\theta}{4}\in\mathbb Q$. Half angle formulae then imply that $\sin\frac{\theta}{2}$ and $\cos\frac{\theta}{2}$ are rational too.
For two points $(\cos\theta,\sin\theta)$ and $(\cos\phi,\sin\phi)$ belonging to $S$, their distance is given by
$$\sqrt{2(1-\cos(\theta-\phi))}=2\left|\sin\left(\frac{\theta}{2}-\frac{\phi}{2}\right)\right|=2\left|\sin\frac{\theta}{2}\cos\frac{\phi}{2}-\sin\frac{\phi}{2}\cos\frac{\theta}{2}\right|\in\mathbb Q.$$
So $S$ satisfies the claim. $\square$
Now we can just pick $N$ points belonging to $S$, scale up by a suitable common denominator and we are done.
A: Pythagorean triples indeed help with finding the solution, and line plus a point provides one example. Another one, in my opinion slightly more elegant, is a circle.
So let us look at points on the unit circle. For each angle $\theta$, the point $(\cos \theta, \sin \theta)$ lies on the unit circle. For some particular $\theta$, both coordinates of this point will be rational – indeed, points of form $\left(\frac{m^2-n^2}{m^2+n^2},\frac{2mn}{m^2+n^2}\right)$ are rational and lie on the unit circle (this is where we use Pythagorean triples). Let $S$ be the set of all such $\theta$.
What is particularly useful about $S$ is that it is closed under addition. Indeed, we can express $\cos(\theta + \phi)$ and $\sin(\theta+\phi)$ by $\cos \theta$, $\sin \theta$, $\cos \phi$ and $\sin \phi$, which are rational numbers whenever $\theta,\phi \in S$. We also have a useful property $\theta \in S \iff -\theta \in S$.
Now the distance between $(\cos \theta, \sin \theta)$ and $(\cos \phi, \sin \phi)$ is the same as the distance between $(\cos (\theta-\phi), \sin(\theta-\phi))$ and $(1,0)$ – this is just rotation by $\phi$. Thus instead of proving "any two points coming from $S$ have rational distance" we can prove "any point coming from $S$ has a rational distance from $(1,0)$". The only issue is that it is not true, but let's do necessary calculations anyway. Distance between $(\cos \theta, \sin \theta)$ and $(1,0)$ is equal to
$$ \sqrt{(\cos \theta - 1)^2 + \sin^2 \theta} = \sqrt{\cos^2 \theta - 2 \cos \theta + 1 + \sin^2 \theta} = \sqrt{2 - 2 \cos \theta}. $$
This does not have to be rational, but if we remember that $\cos \theta = 2 \cos^2 \frac{\theta}{2} - 1$, we can write the expression above as $2 \sqrt{1 - \cos^2 \frac{\theta}{2}}$, which is $2\sin \frac{\theta}{2}$.
Thus, if we take the set of points $(\cos 2 \theta, \sin 2 \theta)$ where $\theta \in S$, then any two points are in rational distance from each other.
To solve your problem, just choose $N$ points from this set, and scale the picture up by the common denominator, so that instead of rational distances we have integer distances.
