Basic Algebra Time
Let's see it Veeky Forums :^)
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