**x**

^{n}+ y^{n}= z^{n}I will show that given any three integers that satisfy this equation, either:

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

or

(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,yare said to becoprimeif and only ifddividesx, ddividesy -> d = 1Two numbers

x,yare said to not becoprimeif and only if there exists a valued > 1such thatddividesx, ddividesy

Now, here's the proof that was promised:

**Lemma: We can reduce any solution to x**

^{n}+ y^{n}= z^{n}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 x

^{n}+ y

^{n}= z

^{n}, 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) z

^{n}= x

^{n}+ y

^{n}= (dx')

^{n}+ (dy')

^{n}

= d

^{n}(x')

^{n}+ d

^{n}(y')

^{n}

= d

^{n}[(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 y

^{n}= z

^{n}- x

^{n}

(4) We can now follow the same reasoning as above.

QED

**Step 2: d**

^{n}divides x^{n}→ 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

**(X**.

^{n},D^{n}) = 1(5) We know that there exists

**k**such that

**x**[Since

^{n}= k * d^{n}**d**divides

^{n}**x**]

^{n}(6) Applying (2), we get

**(cX)**

^{n}= k*(cD)^{n}(7) Which gives us:

**c**

^{n}X^{n}= k * c^{n}D^{n}(8) Dividing

**c**from each side gives:

^{n}**X**

^{n}= D^{n}*k(9) Now it follows that gcd(

**D**) =

^{n},k**1**.

(a) Assume gcd(D(10) So, we can conclude that^{n},k) = a, a > 1

(b) Then,adivides D^{n}and X^{n}[From 8]

(c) But gcd(D^{n},X^{n}) ≠ 1.

(d) But this contradicts (4)

(e) So, we reject our assumption.

**k**is an n-power. [See blog on Infinite Descent for the proof]

(11) Which means that there exists

**u**such that

**u**.

^{n}= k(12) And we get

**D**

^{n}* u^{n}= X^{n}(13) And

**(Du)**

^{n}= X^{n}(14) Implying that

**Du = X**and multiplying by

**c**that

**du=x**.

(15) Which proves that

**d**divides

**x**.

QED

## 12 comments :

At the end of the proof does

(12) And we get Dn * un = Zn

mean

(12) And we get Dn * un = Xn?

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

Thanks very much for noticing this!

-Larry

"(14) Implying that Du = X and multiplying by c that du=X." I think that should instead end with "...by 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.

Thanks for noticing that. I just fixed it.

-Larry

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

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

David

UMMM.. WER DID U GET THE D ONE INSTEAD THE GIVEN IS X,Y,Z ONLY?

SORRY IM NOT A MATHEMATICIAN

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.

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,

QED.

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

Post a Comment