System of Equations

Impossible to solve, proven by Mr Wiles haha :>>>>>)

>>8508873
Link to the proof?

No integer solutions, since [math]x + y + z \equiv x^2 + y^2 + z^2 \mod 2[/math]

>>8508887
But it doesn't have to be integer solutions

>>8508888
Nah, but it's still something. I'm thinking about the general solutions.

>>8508873
But 3 equations with three unknowns should be able to be solved

>>8508888
Okay, it's solvable but I don't think it's pretty:
Write [math]y = 1 - x - z[/math]
Now [math]x^2 + (x+z - 1)^2 + z^2 = 34[/math] and [math]x^3 + (x+z-1)^2 + z^3 = 97[/math]
Set [math]u = x+z, v = xz[/math]
The system becomes [math]2u^2 - 2u - 2v = 34[/math] and [math]u^3 +u^2 - 3uv - 2u = 96[/math]
Substituting v in the second equation, we find a third degree equation in u, which is solvable using Cardan's formulas. That gives us 3 values of u, and hence 3 corresponding values of v, and hence, solving each quadratic equation [math]x^2 - ux + v = 0[/math], we get 6 values of (x,y), and 6 corresponding values of z. But I'm not doing it by hand.

>>8508904
It can be solved.
Solve the first for z => z=1-x-y, substitute into the second and third: Solve the second (quadratic) for x, substitute into the third. That gives you a cubic in y:
4*y^3 - 4*y^2 - 99*y - 93 = 0
That can be solved (it has 3 real roots), but it's ugly. Numerically, the solutions are
-1.02544845449982
-3.855398344452967
5.880846798952788 [*]
Back-substitute into the second and first to get x and z. The third value marked [*] results in complex x, the other two give x, z as
-2.917748692529978, 4.943197147029798
0.5108714460169772, 4.34452689843599

IOW:
x = -2.917748692529978 y = - 1.02544845449982 z = 4.943197147029798
x = 4.943197147029798 y = - 1.02544845449982 z = -2.917748692529977
x = 0.5108714460169772 y = - 3.855398344452967 z = 4.34452689843599
x = 4.34452689843599 y = - 3.855398344452967 z = 0.5108714460169774

>>8508904
Only if the system is linear
[math]X^2 + Y^2 + Z^2 =34[/math] is not linear

>>8508869
I don't know but if a solution exists then at least one number must be bigger than 1 and at least one number must be negative.

There is a simple trick to solve those types of problems. You always have to convert everything into elementary symmetric polynomials.

[math] XY + YZ + ZX = \frac{(X + Y + Z)^2 - (X^2 + Y^2 + Z^2)}{2} = -\frac{33}{2} [/math]
[math] XYZ = \frac{(X^3 + Y^3 + Z^3) + 3 (X + Y + Z) (XY + YZ + ZX) - (X + Y + Z)^3}{3} = \frac{31}{2}[/math]

Now consider the polynomial

[math] p(t) = (t - X) (t - Y) (t - Z) = t^3 - (X + Y + Z) t^2 + (XY + YZ + ZX) t - XYZ = t^3 - t^2 - \frac{33}{2} t - \frac{31}{2}[/math]

Clearly the roots of the polynomial are X, Y and Z so you just have to use the cubic formula and you get some long ugly expressions involving third roots for X, Y and Z.

File: SmartSelectImage_2016-12-01-17-42-33.png (124KB, 1601x547px) Image search: [Google]

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What about other solutions to this?

>>8509014
You win

>>8508873
Get taint slapped

>>8509123
>Get taint slapped

Papa Wiles proved there are no integer solutions and that is true, dummy dumbo hehehehe.

And the integers are the only set of numbers that really exist sooooooooooo..................

>>8508869
SQT

this is one of the very few cases where algebraic geometry is actually useful

i'll work on a proof when i get home unless someone does it first

>>8509251
A proof would probably be pretty helpful

>>8509141
If any integer can pair 1:1 with any decimal number, then this may be true but what about irrationals like π or tau?

>>8508869

x+y+z = 1
x*x + y*y + z*z = 34

x*(x-1) + y*(y-1) + z*(z-1) = 33?

x^3 + y^3 + z^3 = 97

x*x*(x-1) + y*y*(y-1) + z*z*(z-1) = 63?

o = 3.1857
3o^3 = 97
[ n = +- 3.3665
3n^2 = 34
n2 = +- 5.8309
n2^2 = 34 ]

heh nice meme :^)

File: too smarr.gif (27KB, 24x20px) Image search: [Google]

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>>8508869

File: sci_yz_plot.gif (8KB, 555x298px) Image search: [Google]

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I tried to make this into something vaguely visualize-able

Given the following curves:
x^2 = x*x
34 - y^2 - z^2 = (1 - y - z)^2
and
x^3 = (x^2)*x
97 - y^3 - z^3 = (1 - y - z)(34 - y^2 - z^2)

I plugged them into Wolfram Alpha and came out with 6 real combinations of y and z, shown on the graph.

Then, I simply solved for x, giving the following combinations:

(4.958262, -2.868435, -1.089826)
(-1.089826, -2.868435, 4.958262)
(4.958262, -1.089826, -2.868435)
(-2.868435, -1.089826, 4.958262)
(-1.089826, 4.958262, -2.868435)
(-2.868435, 4.958262, -1.089826)

More digits on those truncated numbers are
-1.08982605773566374713135001
-2.8684354534478466425851599
4.9582615111835103897165100

File: image.jpg (982KB, 2956x2246px) Image search: [Google]

982KB, 2956x2246px

>>8510005
This is beautiful anon

>>8508869
did you really mean Y^2 in the third equation?

>>8510020
Thank you, I was afraid I would get chastised for using Wolfram Alpha

would be curious to see if anyone has exact values for the results, even in messy cube root terms like >>8509095 said

>>8509105
Solution is to fix your shitty hand writing

>>8508873
X Y Z need not be integers

so what's the closed form solution?

>>8510021
Yes indeed, if it was Y^3 then this would hardly be a thread. It would be too easy.

>>8510153
i'm a retard please help
what is the solution if it's a Y^3 and not a Y^2?

>>8510038
[math]X=\frac{1}{3} - \frac{(1 - i\sqrt{3}) \sqrt[3]{1138 + 3 i \sqrt{85062}}}{6\sqrt[3]4} - \frac{101 (1 + i\sqrt{3})}{6 \sqrt[3]{2 (1138 + 3 i\sqrt{85062})}}[/math]

[math]Y=\frac{1}{3} - \frac{(1 + i\sqrt{3}) \sqrt[3]{1138 + 3 i\sqrt{85062}}}{6\sqrt[3]{4}} - \frac{101 (1 - i\sqrt{3})}{6 \sqrt[3]{2 (1138 + 3 i \sqrt{85062})}}[/math]

[math]Z== \frac{1}{3} + \frac{\sqrt[3]{1138 + 3 i \sqrt{85062}}}{3\sqrt[3]{4}} + \frac{101}{3\sqrt[3]{2 (1138 + 3 i\sqrt{85062})}}[/math]

File: sci_xz_plot.gif (7KB, 499x298px) Image search: [Google]

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>>8510153
fug :DDD

OK I redid what I did here>>8510005 and found FOUR real combinations of x and z, graph related.

Solving for y, I found the following real solutions

(-2.9177487, -1.025448, 4.9431971)
( 0.5108714, -3.855398, 4.3445269)
( 4.3445269, -3.855398, 0.5108714)
( 4.9431971, -1.025448, -2.9177487)

more digits:
-2.9177486925299777346312773
0.51087144601697742455439351
4.3445268984359898475106573
4.9431971470297980931933608

>>8509014 got it right first

>>8510509
also
-1.0254484544998203585620835
-3.85539834445296727206505081
for the y-values

The intersection of the first two surfaces is a circle, you can parameterize it as

X=1/3 - cos(t)*sqrt(101/6)-sin(t)*sqrt(101/18);
Y=1/3 + cos(t)*sqrt(101/6)-sin(t)*sqrt(101/18);
Z=1/3 +sin(t)*sqrt(101/18)*2;

Putting that into X^3 + Y^3 + Z^3 - 97 =0 gives--1138 = 101 sqrt(202) sin(3 t)
which you can easily solve using arcsin().

The X^3+Y^2+Z^3 form is messier.

>>8508869
{{x -> 2.49725, y -> -5.26913, z -> 3.77188}, {x -> 4.33823,
y -> -3.89612, z -> 0.557888}, {x -> 5.82844, y -> -0.171037,
z -> -4.65741}, {x -> -5.82842, y -> 0.171756, z -> 6.65667}}

File: Untitled.png (21KB, 1152x1080px) Image search: [Google]

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>>8512079
forgot image. there are also some complex solutions if anyone waNTS

File: 1457403416955.png (7KB, 420x420px) Image search: [Google]

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Just plug it into MATLAB, brainlets

>>8512081
>x^2+y^2 = 34

miss a z^2 ?

>>8509016
found the sophomore