For what integer values of $m$ and $n$ is $\frac{4m-n}{n}$ a rational square?

Question

Question for what integer values of $m$ and $n$ with $(m,n)=1$ is $\frac{4m-n}{n}$ a rational square?

Note the motivation for this question is a curiosity i noticed, that the smallest angle of the 3-4-5 triangle is equal to the smallest angle of a triangle whose sides are $\sqrt{5}, \sqrt{5}$ and $\sqrt{2}$.

Answer 1

You are asking for rational points in the parabola $$x^2-4y+1=0.$$ It is easy to find one of such points, say $(0, 1/4)$. Take a line of rational slope $q$ that passes through $(0,1/4)$ and compute the intersection between the line and the parabola.

The line is $y=qx+\frac14$. Put this is the parabola equation to get $x^2-4qx=0$. Since you don't want the point where $x=0$ (a line and a parabola generally intersect at two points) the other point must have $x=4q$. This also shows the $y$ of the new point is $4q^2+\frac14$.

That is, take any rational $q$. Then $(x,y)=\left(4q, 4q^2+\frac 14\right)$ is a solution of our desired equation. Note that in the original problem, $m/n$ is our $y$. A more careful analysis shows these are actually all the solutions.

Answer 2

It should be written in the General form.

$$\frac{4x-y}{y}=z^2$$

$$y=2k$$

$$x=(2t^2+2t+1)k$$

$$z=2t+1$$

$y$ - can only be an even number.

Because the equation $x^2-4y+1=0$ has no solutions in integers.