A question about pythagorean triples Recently, I was given a problem, which was to find two sets of points on the graph $y = x^2$ that have a rational distance from each other. I was then told, if I couldn't find any, to try and prove none exist, and if I could, to find a general solution.
To try and solve the problem, I started with the distance formula, and set it equal to $p\over q$, since the solutions have to be rational: 
$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = {p\over q}$ 
I noted that any point on the graph would be in the form $(x,x^2),$ so I modified the distance formula, and also, squared both sides. 
$(x^2 - x)^2 + (y^2 - y)^2 = ({p\over q})^2$ 
I then rearranged the terms on the left side 
$(x^2 - x)^2 = (x(x - 1))^2 $
$(x(x-1))^2 + (y(y -1))^2 = ({p\over q})^2$
Since this looks like the pythagorean theorem, I focused on trying to find natural number solutions, which would be pythagorean triples that satisfy this equation. I figured all numbers in the triple have to be even, and the ones on the left side have to be able be factored in the form $x \times (x-1)$. I looked at lists of triples, but can't find any to satisfy those two properties. Also, if I make a list of numbers which can be factored into that form, $12, 20, 30, 42$, etc. I can't find any of those numbers in a triple. 
I am stuck here. I have no way to prove if any solution of this type exists or doesn't exist. How would I go about doing that, and if one does exist, how would I find it? 
 A: As P Vanchinathan noted in a comment, you confused the coordinates of the points. Starting again from
$$\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = {p\over q}$$
and now using the correct relationship $y_i=x_i^2$, we get
$$
\sqrt{(x_2-x_1)^2+(x_2^2-x_1^2)^2}=\frac pq\;,\\
(x_2-x_1)^2\left(1+(x_2+x_1)^2\right)=\frac{p^2}{q^2}\;.
$$
If $x_1$ and $x_2$ are rational, the factor $(x_2-x_1)^2$ is a square of a rational number, so we just need to ensure
$$1+(x_2+x_1)^2=\frac{p'^2}{q'^2}$$
with $x_1$ and $x_2$ rational. Setting $x_i=p_i/q'$, we get
$$
q'^2+(p_2+p_1)^2=p'^2\;.
$$
Thus we can take any Pythagorean triple and use one of the legs as $q'$ and divide the other leg up between $p_1$ and $p_2$ in any way. For instance, from the triple $(3,4,5)$ we can get $q'=3$, $p_1=1$, $p_2=3$, corresponding to $x_1=\frac13$ and $x_2=1$, or $q'=4$, $p_1=1$, $p_2=2$, corresponding to $x_1=\frac14$ and $x_2=\frac12$.
A: Despite your already mentioned error, I do not believe your question poser intended you to go through this route. If I understand correctly, your question poser did not necessarily want you to find all points. $f(x)=x^2$ is not one to one, in fact it is an even function so:
$$f(x)=f(-x)$$
So it would be much easier if you try to find a rational distance from $(x,f(x))$ to $(-x,f(x))$.
In which case you have:
$$\sqrt{(x-(-x))^2+(f(x)-f(-x))^2}=\frac{p}{q}$$
But because $f(x)=x^2=f(-x)=(-x)^2=(-1)^2x^2$ we have:
$$\sqrt{(x-(-x))^2+0}=\frac{p}{q}$$
$$|2x|=\frac{p}{q}$$
$$2|x|=\frac{p}{q}$$
The result should be easy to see because $(-x,f(-x))$ and $(x,f(x))$ lie in a straight line. Now all we need is twice $|x|$ to be rational, and because the rationals are closed under multiplication all we need is $|x|$ to be rational.
So if you pick  $x \in \mathbb{Q}$ and with it $-x \in \mathbb{Q}$ you will get solutions to your problem. 
