If we have: $$x^2 + xy + y^2 = 25 $$ $$x^2 + xz + z^2 = 49 $$ $$y^2 + yz + z^2 = 64 $$
How do we calculate $$x + y + z$$
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There are solutions $(x,y,z)=(-5,0,8), \ (5,0,-8)$ having different sums $3,-3$, and two others I found on maple, numerically roughly $(-2.2,-3.5,-5.6)$ with sum around $-11.35$, and the same with all positive signs. So my guess is there's not a clever way to get at the sum, however the thing is at worst quadratic for the third answer.
ADDED: By subtracting equations and factoring, one can get three expressions for $x+y+z$ of the type constant over a difference of two of the coordinates. Setting these equal in pairs one finds that either some two of the varibles are equal (I didn't check that leads nowhere), or else in each case the same relation $8x-13y+5z=0$ results. Solving this for $y$ and plugging into the equations, and messing around a bit, leads to an equation for $z$ which factors as $(z-8)(z+8)(129z^2-4096).$ So besides the two integer solutions there may be others for which $z=\pm (64/\sqrt{169})$.
Going back to the other equations then gives $(x,y,z)=(25,40,64)/\sqrt{129}$ [notation meaning deivide each by the radical] as a solution, as well as its opposite obtained by making all signs negative. Looking at other combinations for values of $x,y$ from the equations (given this $z$) did not produce other solutions, in agreement with what maple found.
coffeemath
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I am positive that the values of x,y and z are complex – imranfat May 22 '13 at 18:47
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@imranfat -- Why don't you at least plug in the integer soloutions $(-5,0,8)$ and its negative, before asserting the values of $x,yz$ are complex?? – coffeemath May 22 '13 at 19:45
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I thought they had to be complex because the quadratic terms aren't factorable over real numbers, but you proved me wrong – imranfat May 22 '13 at 19:49
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You just got a positive "click"... – imranfat May 22 '13 at 19:49
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I saw that term on this site before: What is a "click"? – coffeemath May 22 '13 at 19:51
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Behind your name there is a black number. That goes up as people are favouring your contribution. I call it a click, but that's not official. – imranfat May 23 '13 at 15:40
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How about "upvote" and "reputation", @imranfat? – The Chaz 2.0 Jul 22 '13 at 00:37
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$$x^2 + xy + y^2 = 25 \dots (1)$$ $$x^2 + xz + z^2 = 49 \dots(2)$$ $$y^2 + yz + z^2 = 64 \dots(3)$$
$(2)-(1)$
$x(z-y)+(z+y)(z-y)=24$
$\Rightarrow (x+y+z)(z-y)=24$
Similarly we get ,
$(x+y+z)(y-x)=15$ by $(3)-(2)$
$(x+y+z)(z-x)=39$ by $(3)-(1)$
Clearly Let $1/\lambda=(x+y+z)\ne 0$
Then we have,
$(z-y)=24\lambda$
$(y-x)=15\lambda$
$\Rightarrow x+z-2y=9\lambda$
$\Rightarrow 3y=1/\lambda+9\lambda\Rightarrow y=1/3\lambda+3\lambda$
Now solve for x and put in the first equation to find the value of $\lambda$.
Abhra Abir Kundu
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I get a system of infinite solution when eliminating z. I end up with two equations of y-x = 15lambda So what am I doing wrong? – imranfat May 22 '13 at 19:00
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I cant understand how u can get infinite solutions for $\lambda$ because after putting the values of x and y in (1) you will get a quadratic possibly in $\lambda^2$ – Abhra Abir Kundu May 22 '13 at 19:08
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That's because I subtracted in the nature of I-II then II-III and then I-III and that's not right. Obviously that results in a dependent situation. It makes sence to get 2 solutions. – imranfat May 22 '13 at 19:42
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@imranfat ... Just added a section in my answer looking at finding the answers. – coffeemath May 22 '13 at 21:19