- #1
Loren Booda
- 3,125
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Does the equation
a3 + b3 = c3
(where a, b and c are constants) have any general geometrical significance?
a3 + b3 = c3
(where a, b and c are constants) have any general geometrical significance?
Loren Booda said:Does the equation
a3 + b3 = c3
(where a, b and c are constants) have any general geometrical significance?
John Creighto said:What about if we give a triangle depth and do something like the Pythagoras theorem but use area instead of length?
tiny-tim said:No way!
I don't think cubes ever arise in geometry.
Maybe in calculus, and in some branches of physics, but not in geometry.
tiny-tim said:No way!
I don't think cubes ever arise in geometry.
tiny-tim said:oh, those cubes! ™
Bananaman3579 said:I'm still not seeing why a cube root number wouldn't cancel out the cube. I know it has no practical application but that equation i gave would work. what you left out in your proof is the cube root so it should be the (cube root of 5) to the third power. This also got me thinking about the ratio of three dementional object sides to each other, does anyone know some equations along with their application in regards to shape.
tiny-tim said:… or Russian dolls! …
mathwonk said:have you heard of elliptic curves? there is a lot of beautiful geometry associated with cubic plane curves like this.
Alex48674 said:Sorry Banana I miss read you a bit:
But I figured out where it is wrong
So basically what it is is that the problem is a^3+b^3=c^3
but you changed it to cube root a^3 +cube rootb^3 =cube rootc^3
which is equal to a+b=c, so if what you say is right you would be able to use any terms for this second equation and it would fit the third
so let's say 1+1=2
then plug in you get 1+1=8... doesn't work. What you have done is changed the problem completely, you can not justify taking the cubic roots of all the terms and still call it the same problem. So in short it is not a solution. In fact there is no solution, it is impossible to find one.
Mute said:As far as I can tell Bananaman was simply trying to provide a set of numbers that solved the equation a^3 + b^3 = c^3. He never made any reference to trying to find natural numbers that solved the problem. Hence, for any x, y and z such that x = a^3, y=b^3 and z=c^3, the equation is solved. You've been assuming he's been trying to find natural number solutions and hence have been telling him there is no solution, but if he's just trying to find a triple of real numbers that satisfies the equation, then he's done that and telling him there's no solution is bound to just be confusing.
Bananaman3579 said:Yeah I might be in a little over my head here but all I was finding was a set of numbers that would make that equation true, NOT a solution that would work infinitly. basically what I'm saying is that you can take any two numbers, take th cube root of them and plug that into A and B. thus making the answer to the equation as the addition of the first two original numbers.
A^3+B^3=C^3
A= cube root 5
B= cube root 9
C= cube root 14
plug them in and the equation is true. CALCULATOR WILL PROVE ALL...or explode
Mute said:Yes, that's true, but the reason the people found the problem interesting was that Fermat supposedly had a theorem (which never got written down) that showed that the equation
a^n + b^n = c^n
has no solutions such that a, b and c are Natural Numbers (integers greater than zero) for n > 2. There are of course an infinite number of possible triples (a,b,c) that do solve the equation: if you set c = 5, for example, and plot y = (5^3-x^3)^(1/3) you'll get a curve, points along which will satisfy x^3 + y^3 = 5^3, but not both of x and y will be natural numbers, by the theorem (which was not actually proven until recently by Wiles).
Bananaman3579 said:wasnt there an error in Wiles work too?
Loren Booda said:Thank you, mathwonk.
I believe elliptic curves may be near impossible to solve analytically.
The Wiki article on elliptic curves quotes my old professor, Serge Lang.