Wednesday, May 04, 2005

CoPrime Numbers: xn + yn = zn

This blog is about the equation:
xn + yn = zn

I will show that given any three integers that satisfy this equation, either:

(a) all three of them are coprime with each other


(b) it is possible to cancel out common components and derive three numbers that are coprime.

Two numbers are coprime if they do not share any common divisors.

Definition 1: Coprime

Two numbers x,y are said to be coprime if and only if d divides x, d divides y -> d = 1

Two numbers x,y are said to not be coprime if and only if there exists a value d > 1 such that d divides x, d divides y

Now, here's the proof that was promised:

Lemma: We can reduce any solution to xn + yn = zn to a form where x,y,z are coprime.

To prove this, we will need to prove two things:

(1) If a factor divides any two values of this equation, then the n-power of it divides the n-power of the third value.

(2) If an n-power of a factor divides the n-power of a value, then the factor divides the value itself.

Step 1: For xn + yn = zn, the n-power of any common factor of two divides the n-power of the third.

Case I: Let's assume d divides x, d divides y

(1) There exists x', y' such that: x = d(x'), y = d(y')
(2) zn = xn + yn = (dx')n + (dy')n
= dn(x')n + dn(y')n
= dn[(x')n + (y')n]

Case II: Let's assume d divides z and d divides x or d divides y

(1) Let's assume d divides x (the same argument will work for y)
(2) There exists x', z' such that: x = d(x'), z = d(z')
(3) We now say yn = zn - xn
(4) We can now follow the same reasoning as above.


Step 2: dn divides xn → d divides x

(1) Let c be the greatest common denominator (gcd) for d,x.
(2) Let D = d / c, X = x / c.
(3) Now the gcd of (X,D) = 1. [See here for the explanation]
(4) So, the gcd of (Xn,Dn) = 1.
(5) We know that there exists k such that xn = k * dn [Since dn divides xn ]
(6) Applying (2), we get (cX)n = k*(cD)n
(7) Which gives us: cnXn = k * cnDn
(8) Dividing cn from each side gives: Xn = Dn*k
(9) Now it follows that gcd(Dn,k) = 1.
(a) Assume gcd(Dn,k) = a, a > 1
(b) Then, a divides Dn and Xn [From 8]
(c) But gcd(Dn,Xn) ≠ 1.
(d) But this contradicts (4)
(e) So, we reject our assumption.
(10) So, we can conclude that k is an n-power. [See blog on Infinite Descent for the proof]
(11) Which means that there exists u such that un = k.
(12) And we get Dn * un = Xn
(13) And (Du)n = Xn
(14) Implying that Du = X and multiplying by c that du=x.
(15) Which proves that d divides x.



Anonymous said...

At the end of the proof does
(12) And we get Dn * un = Zn
(12) And we get Dn * un = Xn?

Larry Freeman said...

Yes, you are right. I just made the change.

Thanks very much for noticing this!


Anonymous said...

"(14) Implying that Du = X and multiplying by c that du=X." I think that should instead end with " c that du=x." It's nothing big... I'm just hoping others dont get confused by it. Also, thanks, I found your site very helpful.

Larry Freeman said...

Thanks for noticing that. I just fixed it.

David K said...

I think that the proof can be simplified: From (8) we have that D^n divides X^n. But gcd(X^n,D^n)=1. Hence D^n~1 and also D~1. Now from (2) we have c=d.

David said...

Explanation: I meant simplification of step 2 of your proof.

don said...



mattsucks said...

if u define the equal square of zero vontax then u get the same ammount of Du = X so D^n + D~1 equals the 1st du=6 in the multiplied by the x* method.

Anders H said...
This comment has been removed by the author.
Anders H said...

Intuitively, step 2 is trivial. Here is a very simple proof of step 2. Is it flawed?

d^n divides x^n =>
There is an integer
e = (x^n)/(d^n) = (x/d)^n
(by common arithmetic laws).
Apparently there is an integer
x' = x/d such that (x')^n = e.
Apparently (since x/d is integer),
d divides x,

이영지 said...
This comment has been removed by the author.
Unknown said...

Surely if D^n×k=X^n then gcd(D^n,X^n)=D^n?