x

^{n}+ y

^{n}= z

^{n}

We can assume that x,y,z are relatively prime. See here for the details.

It turns out that we can make the same statement about all Euclidean Integers. For those not familiar with Euclidean Integers, please review here.

Lemma 1: For Euclidean Integers, relatively prime divisors of n-powers are themselves n-powers.

This theorem says that if

**gcd(v,w) = 1**and

**vw = z**

^{n}Then, there exists

**x,y**such that

**v = x**

^{n}, w = y^{n}(1) So, we start with

**gcd(v,w) = 1, vw = z**

^{n}(2) Assume that

**v**is not equal to any number

**x**

^{n}(3)

**v ≠ 1**since 1 is an

**x**

^{n}power(4) Now,

**v**is divisible by a prime number

**p**. [Fundamental Theorem of Arithmetic for Euclidean Integers]

(5) So, there exists

**k**such that

**v = pk**

(6)

**p**divides

**z**since

**z**[By applying Euclid's Lemma for Euclidean Integers]

^{n}= vw = pkw(7) So, there exists

**m**such that

**z=pm**

(8) So,

**z**

^{n}= vw = pkw = (pm)^{n}= p^{n}m^{n}(9) Dividing

**p**from both sides gives us:

**kw = p**

^{(n-1)}m^{n}(10) From Euclid's Lemma,

**p**divides

**k**or

**w**.

(11) It can't divide

**w**since it already divides

**v**and

**gcd(v,w)=1**. Therefore, it divides

**k**

(12) We can apply this same argument for each

**p**in

**p**

^{(n-1)}(13) So, we can conclude that

**p**divides

^{(n-1)}**k**.

(14) So, there exists

**V**such that

**k = p**

^{(n-1)}*V(15) So,

**kw = p**

^{(n-1)}m^{n}= p^{(n-1)}*V*w(16) Dividing

**p**from both sides gives us:

^{(n-1)}**vW = m**

^{n}(17) Now,

**gcd(V,w)=1**since

**V**is a divisor of

**v**and

**gcd(v,w) = 1**

(18) Likewise,

**V**cannot be an n-power. If it were, then

**v = p**would make

^{n}V**v**an n-power which goes against our assumption.

(19) Finally,

**V**is less than

**v**since

**p**.

^{(n-1)}> 1(20) Thus, we have a contradiction by infinite descent.

QED

Lemma 2: Given x

^{n}+ y

^{n}= z

^{n}and x,y,z are Euclidean Integers, we can assume that x,y,z are relatively prime.

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 proof.]

(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. [By the lemma above]

(11) Which means that there exists

**u**such that

**u**.

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

**D**

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

**(Du)**

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

**Du = Z**and multiplying by

**c**that

**du=z**.

(15) Which proves that

**d**divides

**z**.

QED