Diophantine equation in rationals $4 k + m^2=n^2$

Question

I am trying to solve the Diophantine equation

$$4 k + m^2 = n^2$$

in rationals.

This looks very simple, but I stuck with this. I have represented this in terms of integers, but again no success:

$$k_2 n_2^2 m_1^2 + 4 k_1 n_2^2 m_2^2= k_2 n_1^2 m_2^2$$

where $k=\frac{k_1}{k_2},m=\frac{m_1}{m_2},n=\frac{n_1}{n_2}$.

EDITED:

I am trying to find such $k$ and $m$ that

$$4k+m^2$$ is square of rational number.

EDITED 2: The problem I am working on is finding of general solution for this equation.

Answer 1

If that you want is some like a parametrization, you can take $ n = \frac{a + b}{2} $, $ m = \frac{a - b}{2} $. Then $$ k = \frac{n^2 - m^2}{4} = \frac{ \left( \frac{a + b}{2} \right)^2 -\left( \frac{a - b}{2} \right)^2 }{4} = ab / 4 $$ So, all the solutions are given by $$ \begin{cases} k = ab / 4 \\ n = \frac{a + b}{2} \\ m = \frac{a - b}{2} \end{cases}, \ \ a, b \in \mathbb{Q} $$

Answer 2

There are clearly infinitely many solutions. For every rationals $m$ and $n$ just take $k=\frac{n^2-m^2}4$.

Looking for solutions with fixed $k$ is a bit more interesting, but still easy.

You can write $4k=(m+n)(m-n)$ and solve, for example, the system $$\left\{\begin{array}{rcl}m+n&=&4\\m-n&=&k\end{array}\right.$$ You can choose for the RHS's any expressions whose product is $4k$.