Why two given equations cannot both be true (fermat's for n=3)

  • Context: Graduate 
  • Thread starter Thread starter drd0013
  • Start date Start date
Click For Summary
SUMMARY

The discussion centers on proving that the equations (x+y)=integer3 and (x2-x*y+y2)=integer3 cannot both hold true simultaneously for integer values of x and y, specifically in the context of Fermat's Last Theorem for n=3. The user attempts to demonstrate this by factoring x2+y2 into (x+y) and (x*y+(x-y)2), leading to the conclusion that both expressions must be cubed integers for x3+y3=z3. However, examples provided show that while each term can be a cubed integer, they cannot both be cubed integers for the same x and y values.

PREREQUISITES
  • Understanding of Fermat's Last Theorem, specifically for n=3
  • Basic algebraic manipulation and factorization techniques
  • Knowledge of integer properties and cubed numbers
  • Familiarity with proof techniques, such as modus tollens
NEXT STEPS
  • Research the implications of Fermat's Last Theorem for n=3
  • Study algebraic identities related to sums of cubes
  • Explore proof techniques in number theory, particularly for integer equations
  • Investigate existing proofs or discussions on the impossibility of simultaneous cubed integers
USEFUL FOR

Mathematicians, students studying number theory, and anyone interested in the complexities of Fermat's Last Theorem and integer equations.

drd0013
Messages
2
Reaction score
0
Okay, so I'm trying to figure something out.
Why can (x+y)=integer^3 not be true while
(x^2-x*y+y^2)=integer^3 is true?
where x and y are also integers. The integer to the right is just an undefined integer.
I'm trying to prove Fermat's last theorem x^n+y^n=z^n for n=3.
I was able to define two factors of x^2+y^2 as:
(x+y) and (x*y+(x-y)^2)) which reduces to (x^2-x*y+y^2).
Also, z^3 must be the product of cubed integers. So, (x+y) must itself be a cubed integer as well as (x^2-x*y+y^2) in order for x^3+y^3 to be = to z^3. (Which we know it can't). However, I don't know enough maths to prove that (x+y) cannot be equal to a cubed integer while (x^2-x*y+y^2) is.

Also, is this proof a novel one? I was just doing it for garbages and giggles.
Thanks to any who may be able to help
 
Mathematics news on Phys.org
(x+y)*(x^2 -x*y +y^2) = z^3
the problem is not that x+y can't be a cubed value, it can, or that x^2 -x*y +y^2 can't be a cubed value, it can. it's proving that there is no case where both are cubed values for the same x,y.

let's take an example. 10^3 = 1000.
10^3 = 2^3 *5^3
so ether x+y = 8, or x+y = 125.
let;s assume the smaller one. which means x^2 -x*y +y^2 needs to equal 125.
x = 3 y= 5
3^2 - 3*5 +5^2 = 9 -15 +25 = 19; no where close to 125.
 
Last edited:
Yeah, I'm just not sure how (if it can be) to go about showing that. I'm trying to prove that the first and second terms can't both be equal to cubed integers at the same time. But I am not sure how to set up or solve a proof of that. I'd assume modus tolens of some kind but don't really know
 
Last edited:

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K