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If We have $x+y=4 $, $ x\cdot y=-1$ and $x>y$ than $9 x^2+15 y^2=\;\;? $

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  • $\begingroup$ We had a similar problem a few days ago and Bill Dubuque suggested using Gauss' algorithm (see math.stackexchange.com/questions/652252/…). I tried and for me it seems to fail for this example, probably I did something wrong. It breaks after second step when only $6y^2$ is left with lex-order $(0,2)$ which means one would have to caculate $6y^2-6(x+y)^{0-2}(xy)^2$. I don't think this leads to the solution here. Otherwise this would make a very good answer. $\endgroup$
    – Piwi
    Feb 1 '14 at 15:03
  • $\begingroup$ This problem can't have a unique solution. Suppose $x=x_0,y=y_0$ is a solution, then so is $x=y_0,y=x_0$. So, unless $x_0=y_0$, there will be two solutions. $\endgroup$
    – dexter04
    Feb 1 '14 at 15:09
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    $\begingroup$ @dexter04 $x > y$, $xy = -1$ ensures a unique solution. $x$ must necessarily be positive, $y$ negative. See my answer for details. $\endgroup$
    – amWhy
    Feb 1 '14 at 15:11
  • $\begingroup$ oops. sorry i overlooked the fact. $\endgroup$
    – dexter04
    Feb 1 '14 at 15:12
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    $\begingroup$ @Piwi: This problem differs from the problem you are refering to. $9 x^2+15 y^2$ is not a symmetric polynomial. $\endgroup$
    – miracle173
    Feb 1 '14 at 15:53
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It's doable algorithmically by symmetric-antisymmetric decomposition and Gauss's algorithm.

$$\begin{eqnarray} f(x,y) &\,=\,& \overbrace{\frac{1}2 (f(x,y)+f(y,x))}^{\large \rm symmetric\ part} \,+\, \overbrace{\frac{1}2(f(x,y)-f(y,x))}^{\large \rm \color{#c00}{anti}symmetric\ part}\\ &=& 12(x^2+y^2)\, +\, 3(y^2-x^2)\\ &=& 12(\color{#0a0}{\color{#c0a0}{x^2+y^2}})\, -\, 3(\color{#c00}{x-y})(\color{#0a0}{x+y})\end{eqnarray}$$

Then, applying Gauss's algorithm (or inspection) to the $\rm\color{#0a0}{symmetric}$ polynomials and also to the (symmetric!) square of the $\rm\color{#c00}{anti}$symmetric} polynomial, we quickly rewrite these polynomials as polynomials in the elementary symmetric polynomials $\,x+y\,$ and $\,xy,\,$ namely

$$\begin{eqnarray} \color{#0a0}{x^2+y^2} &\,=\,& (x+y)^2 - 2(xy)\\ (\color{#c00}{x-y})^2 &=& (x+y)^2 - 4(xy)\end{eqnarray}$$

Since $\,x-y > 0\,$ we know $\, x-y = \sqrt{(x-y)^2},\,$ so computing these values in terms of the known values of the elementary symmetric polynomials $\,x+y\,$ and $\,xy\,$ then substituting these values into the first equation immediately yields the value of $\,f(x,y).\,$ This method works generally.

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  • $\begingroup$ Thanks! I tried to "symmetrize" inutitively but failed gloriously. Small questions considering the generality: (1) What to do if you don't have $y<x$? Start tracking the different cases? (2) Does taking the square of the resulting antisymmetric always make it symmetric? (Sorry, probably easy to see. I don't.) $\endgroup$
    – Piwi
    Feb 1 '14 at 23:52
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    $\begingroup$ @Piwi If you know the sign of the antisymmetric part $\,y-x\,$ then you can choose the correct square root to take after computing the value of its symmetric square (else the answer involves an unknown sign). Secondly, if $f$ is antisymmetric, i.e. $\,f(y,x) = -f(x,y),\,$ then squaring that equality shows that $f^2$ is symmetric. $\endgroup$ Feb 2 '14 at 0:03
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Consider the fact that $9x^2+15y^2 = 12(x^2+y^2)+3(y^2-x^2)$. Now, $x^2+y^2 = (x+y)^2-2xy$ and $y-x = -\sqrt{(x+y)^2-4xy}$. The $-$sign is taken as $y<x$. So, $y^2-x^2 = (y-x)(y+x) $ can be calculated.

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  • $\begingroup$ Good answer of an odd Question: $+1$ $\endgroup$ Feb 1 '14 at 15:26
  • $\begingroup$ @lab This is a special case of an algorithm that works generally - see my answer. $\endgroup$ Feb 1 '14 at 19:24
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$$\begin{eqnarray} 9x^2+15y^2&=\\ 9(x^2+y^2)-6y(4-y)+24y&=\\9((x+y)^2-2xy)-6xy+24y&=\\ 9(4^2+2)-6(-1)+24y&=\\ 168+24y& \end{eqnarray}$$

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