Diophantine-like Equation

Question

Is there a neat method for solving equations of the form: $$\frac {a^2}{x^2}-\frac {b^2}{(1-x)^2}=c^2$$ where $0<x<1$, by exploiting the symmetry in instead of expanding it as a fully-blown quartic?

Perhaps one could use the substitution $y=\frac 12-x$, but it doesn't seem to go far.

Additional Note: The original equation has $$\begin{align} a^2&=u^2\mu^2\\ b^2&=v^2(1-\mu)^2\\ c^2&=\frac {v^2-u^2}{\gamma^2}\end{align}$$ where $0<\mu <1$. Note sure if this additional information helps.

What restrictions (if any) are there on the values of $a, b, c$, and $x$? — paw88789, Sep 09 '19 at 16:25
@paw88789 have edited to specify that $0<x<1$, and added some notes on the constants. — Hypergeometricx, Sep 09 '19 at 16:31
If $0<x<1$, why have you tagged this "diophantine equations"? Are you looking for solutions in rational $x$? — Adam Bailey, Sep 09 '19 at 21:19
@AdamBailey - There isn't a tag for diophantine-like, and perhaps even that is not suitable. Please change the tag to a more appropriate one. — Hypergeometricx, Sep 11 '19 at 16:49

score 2 · Answer 1 · edited Sep 10 '19 at 05:32

2

Above equation shown below:

$$\frac {a^2}{x^2}-\frac {b^2}{(1-x)^2}=c^2\tag1$$

Equation $(1)$ is actually a hyperbolic curve in disguise.

Taking, $x=(\cos p)^2$ we get $(1-x)=(\sin p)^2$.

For p=30 degree's we get, $(\sin p)^2 =\frac14$ &

$x=(\cos p)^2=\frac34$

Thus we get equation:

$$(4a)^2=(12b)^2+(3c)^2\tag2$$ Equation $(2)$ is satisfied at,

$(a,b,c)=[((5/4),(1/4),(4/3)]$

Hence solution for equation $(1)$:

$(a,b,c,x)=[((5/4),(1/4),(4/3),(3/4)]$

edited Sep 10 '19 at 05:32

José Carlos Santos

427,504

answered Sep 10 '19 at 05:20

Sam

31

Thanks for your solution (+1). For the original question, $a,b,c$ are given, so the objective is to express $x$ in terms of $a,b,c$. However, a closed-form solution might not exist for the general case. – Hypergeometricx Sep 11 '19 at 16:50

Diophantine-like Equation

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