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Can the system \begin{align} x+y& =a\\ x^2+b& =z^2\\ y^2+c&=w^2\\ dzy&=wx \end{align} (with $x,y,z,w,a,b,c,d >0$) be solved for $x, y, w$ and $z$ by hand without cubics (which I could solve... but yikes) or quartics, and using only elementary functions? I was challenged to find a simple solution but after alot of manipulating I'm starting to think it's not possible. I know that once you solve for one variable you can use one of the first three equations to isolate another easily with only roots and etc., and then do that again with the variable you just found until you have them all. The trouble is in getting just one of these variables.

I've tried squaring both sides of the fourth equation and writing in terms of x and y, then using the first equation to get just x, by that left me with a bad quartic. I also though because of the elegance of the second and third equation I'd try some hyperbolic trig subs with two auxilarily variables but that didn't get anything new either. Neither did adding some of these equations, as I assume this is because you're using inputs and information from the same two equations.

I don't mind if Galois theory or anything is used in the answer, but I probably won't understand it (at least, not yet ;)).

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    $\begingroup$ There is no general method for this kind of problem. and no particular symmetry here I can see. Given that you were "challenged to solve it" there may be some underlying trick. The source of the problem might help. $\endgroup$ – Ethan Bolker May 4 at 11:16
  • $\begingroup$ No condition on $x,y,z,w,a,b,c,d$ ? (positivity ...) $\endgroup$ – Jean Marie May 4 at 12:28
  • $\begingroup$ Small observation: if $(x, y, z, w)$ is a solution, so is $(x, y, -z, -w)$. Replacing $w$ with, say, $u = w/z$ might therefore help a little bit. There might be some other symmetries hiding in there as well, but I haven't noticed them yet. :) $\endgroup$ – John Hughes May 4 at 12:33
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    $\begingroup$ As @Ethan Bolker said, it could be of some help to know why you are interested in this issue. In particular, the last equation looks rather un-homogeneous compared to the other ones. $\endgroup$ – Jean Marie May 4 at 16:13
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    $\begingroup$ Your question puzzles me - there's nothing to name, other than to say that this is a complicated nonlinear system of equations. @JeanMarie has gone a long way toward helping you with a numerical solution. $\endgroup$ – Ethan Bolker May 5 at 22:18
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As you didn't show your results, I'll redo the computation.

Let us substitue the RHS of the two middle equations in the fourth, squared:

$$d^2(x^2+b)y^2=(y^2+c)x^2,$$ or

$$(d^2b-x^2)y^2=(c-d^2)x^2.$$

Then, from the first equation,

$$y^2=(a-x)^2$$

and finally, eliminating $y$,

$$(c-d^2)x^2=(d^2b-x^2)(a-x)^2.$$

This confirms that the problem is of the quartic degree and has an analytical solution, but the formulas are terrible and hardly manageable by hand.

And as the polynomial has three independent coefficients, there is no reason that any simplification occurs and you won't find any simpler analytical resolution (otherwise, you would revolutionize the Galoi's theory).


The other option is to resort to numerical methods, but whether this is feasible by hand is unsure, and it is only doable for particular values of the parameters $a,b,c,d$.

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  • $\begingroup$ I suppose this is true for both x and y, but also w and z because they can be written in terms of x and y, therefore solving for either w or z would somehow also be equivalent to solving for x and y which can't be done using the given methods. ? $\endgroup$ – Benjamin Thoburn May 6 at 21:44
  • $\begingroup$ @BenjaminThoburn: " which can't be done using the given methods", what do you mean ? $\endgroup$ – Yves Daoust May 6 at 22:15
  • $\begingroup$ Using methods that express solutions in terms of elementry functions, but not third roots or higher. $\endgroup$ – Benjamin Thoburn May 7 at 8:20
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Let us reduce the issue to the obtention of a root $x_0$ of a certain equation $f(x)=0$ (see figure below).

Knowing that your issue is oriented towards applications, a final step is to use a numerical method such as dichotomy (or Newton's method) to find an approximation to $x_0 \in (0,a)$ ; knowing $x_0$, one will get the unique values of $y$, $z$ and $w$. We will leave this numerical solution up to you.

Let us explain how one arrives at this function $f$.

The different variables $y,z,w$ in system :

$$\begin{cases} x+y& =&a\\ x^2+b& =&z^2\\ y^2+c&=&w^2\\ dzy&=&wx \end{cases}$$

can be progressively transformed into expressions involving $x$ only (and parameters $a,b,c,d$).

$y,z$ and $w$ can be expressed as functions of $x$ in the following way :

$$y=a-x, z=\sqrt{x^2+b}, w=\sqrt{y^2+c}=\sqrt{(a-x)^2+c}$$

(no $\pm$ sign in front of the square roots because $z$ and $w$ are positive) ; inserting these expressions into the fourth equation gives the constraint :

$$d\sqrt{x^2+b}(a-x)=x\sqrt{(a-x)^2+c}$$

which is equivalent, under the condition $a>x$, to :

$$d^2(x^2+b)(a-x)^2=x^2((a-x)^2+c)$$

or

$$d^2\dfrac{x^2+b}{x^2}=\dfrac{(a-x)^2+c}{(a-x)^2} \ \iff$$

$$d^2(1+\dfrac{b}{x^2})=1+\dfrac{c}{(a-x)^2} \ \iff $$

$$\underbrace{(d^2-1)+\dfrac{bd^2}{x^2}-\dfrac{c}{(a-x)^2}}_{f(x)}=0.\tag{1}$$

As

$$f'(x)=-\dfrac{bd^2}{x^3}-\dfrac{c}{(a-x)^3}<0 \ \ \text{for} \ \ x \in (0,a), $$

the curve of $f$ is decreasing on $(0,a)$ from $+\infty$ to $-\infty$, thus with a unique root in this interval. See the graphical representation in the example below, representative of the general case. We don't consider the other branch because $x>a$ isn't allowed.

enter image description here

Fig. 1 : Graphical representation of $f$ (Case $a=5;b=2;c=3;d=2$) with solution $x_0 \approx 4.07186$. Only the left branch is to be considered, because of condition $y=x-a>0$.

Remark : $f(x)=0$ is equivalent to a 4th degree equation :

$$(d^2-1)x^2(a-x)^2+bd^2(a-x)^2-cx^2 \ = \ 0$$

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    $\begingroup$ Many thanks, I wouldn't have thought of this approach myself. In case you you were interested the original values were a=2, b=3.24, c=4, and d=1.33. I was able to find all the (16) roots over complex numbers using Wolfram. Anyways, I thought exact answers in the general case would be more interesting (because I'm a mathematician!, and also therefore not worried about the real world). $\endgroup$ – Benjamin Thoburn May 6 at 21:40
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    $\begingroup$ I am surprised you find 16 roots (see the remark I just added). With the values you have given, I find only 4 roots $x =1.1286346$ (the only convenient root) $x=3.8625404, x= - 0.4955875 \pm 2.567820 i $. $\endgroup$ – Jean Marie May 6 at 22:58
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    $\begingroup$ that was for all four variables in the original system (x,y,w,z). $\endgroup$ – Benjamin Thoburn May 7 at 8:17

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