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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.

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  • $\begingroup$ Do you want to solve it for $k$, $n$ and $m$? Or is $k$ a parameter? $\endgroup$ – ajotatxe Dec 30 '17 at 21:36
  • $\begingroup$ I am looking for triples $k,n,m$. $\endgroup$ – Gevorg Hmayakyan Dec 30 '17 at 21:36
  • $\begingroup$ Correct me if I am wrong, but I thought that when you say "Diophantine equation", the understanding is that you are looking for integer solutions. Is that correct? $\endgroup$ – Manolito Pérez Dec 30 '17 at 21:44
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    $\begingroup$ Actually they are equivalent in general. $\endgroup$ – Gevorg Hmayakyan Dec 30 '17 at 21:48
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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 $$ \begin{cases} k = ab / 4 \\ n = \frac{a + b}{2} \\ m = \frac{a - b}{2} \end{cases}, \ \ a, b \in \mathbb{Q} $$

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  • $\begingroup$ Many thanks. Just wondering if this gives all the solutions? $\endgroup$ – Gevorg Hmayakyan Dec 30 '17 at 21:47
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    $\begingroup$ Yes. If you have $ m,n,k $ such that $ 4k + m^2 = n^2 $, then solve the system $ n = \frac{a + b}{2} $, $ m = \frac{a - b}{2} $ for $ a, b $ (it is easy to see that the solution are rationals number too). Then the equality $ 4k + m^2 = n^2 $ implies $ k = ab / 4 $. $\endgroup$ – M159 Dec 30 '17 at 21:51
  • $\begingroup$ This is overkill, since there is a trivial parametrization of all solutions - given $m,n$ there is exactly one $k$. $\endgroup$ – Thomas Andrews Dec 30 '17 at 22:30
  • $\begingroup$ The question is to find such $k$ and $m$ that the $4k+m^2$ is square. For this we need some parametrization. This is the same complexity task as pythagorean triples or any other 2-order diophantine equation. $\endgroup$ – Gevorg Hmayakyan Dec 31 '17 at 19:17
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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 $4k$.

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  • $\begingroup$ Sorry my fault: Corrected the question $\endgroup$ – Gevorg Hmayakyan Dec 30 '17 at 21:39
  • $\begingroup$ Many thanks, I am just looking for general solution. $\endgroup$ – Gevorg Hmayakyan Dec 30 '17 at 21:48

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