System of Equations

Basic Algebra Time

Let's see it Veeky Forums :^)

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

Link to the proof?

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

But it doesn't have to be integer solutions

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

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

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.

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

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

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.

What about other solutions to this?

You win

Get taint slapped

>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..................

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

A proof would probably be pretty helpful

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

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 :^)

...

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

This is beautiful user

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

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 said

Solution is to fix your shitty hand writing

X Y Z need not be integers

so what's the closed form solution?

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

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

[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]

fug :DDD

OK I redid what I did here 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

got it right first

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.

{{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}}

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

Just plug it into MATLAB, brainlets

>x^2+y^2 = 34

miss a z^2 ?

found the sophomore