Method of Desboves

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Method of Desboves

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#1 h Oct 29, 2012, 8:28 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

From the attached picture . Do anyone know about method of Desboves ?

This problem is in CRUXv27n2 "Murray Klamklin Quickies Problem"

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tenniskidperson3

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tenniskidperson3

#2 h Oct 31, 2012, 3:22 AM • 3 Y

Y by MinatoF, Adventure10, Mango247

I have never heard of the Method of Desboves, but guessing from its application in the problem, I suppose this may be what they're talking about. You have a quadratic, and you want to get a linear equation in some variable. First you find a solution $(x_0, y_0, w_0)$ (possibly with more variables, depending on how many there are in the equation). Then you $w=w_0r$ , $y=y_0r+y_1$ , $x=x_0r+x_1$ , and so on. You let all variables but one be shifts of multiples of a rational variable $r$ ; presumably the one exception is to have fewer unnecessary variables in the end equation. Then the coefficients of $r^2$ on each side are the same, because $(x_0, y_0, w_0)$ is a solution and the coefficient of $r^2$ on each side is exactly what you get when you plug in $x_0, y_0, w_0$ for $x, y, w$ on each side. So the $r^2$ 's cancel and you're left with a linear equation in $r$ . You can solve it, and since the equation in $x, y, w$ is homogeneous, you can multiply all three by a constant to get integers for each.

Clearly this extends to more than 3 variables. You can use the same method to solve $xy+yz+zx=w^2$ ; take $x_0=y_0=1$ and $z_0=4$ and $w_0=3$ . You get $x=r+a, y=r+b, z=4r+c, w=3r$ . The equation reduces to

$(r+a)(r+b)+(r+b)(4r+c)+(4r+c)(r+a)=9r^2$ , or $9r^2+r(a+b+4b+c+4a+c)+(ab+bc+ca)=9r^2$ .

So $r=-\frac{ab+bc+ca}{5a+5b+2c}$ and $x=(5a+5b+2c)a-(ab+bc+ca)$ , $y=(5a+5b+2c)b-(ab+bc+ca)$ ,

$z=(5a+5b+2c)c-4(ab+bc+ca)$ , and $w=-3(ab+bc+ca)$ .

We can multiply these by constants too, to get more solutions. I do not know if this gives all solutions, and it very well might not. You may have to vary $a, b, c$ through rationals to get all solutions.

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