xn + yn = zn
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 = zn
Then, there exists x,y such that v = xn, w = yn
(1) So, we start with gcd(v,w) = 1, vw = zn
(2) Assume that v is not equal to any number xn
(3) v ≠ 1 since 1 is an xn 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 zn = vw = pkw [By applying Euclid's Lemma for Euclidean Integers]
(7) So, there exists m such that z=pm
(8) So, zn = vw = pkw = (pm)n = pnmn
(9) Dividing p from both sides gives us:
kw = p(n-1)mn
(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(n-1) divides k.
(14) So, there exists V such that k = p(n-1)*V
(15) So, kw = p(n-1)mn = p(n-1)*V*w
(16) Dividing p(n-1) from both sides gives us:
vW = mn
(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 = pnV would make 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.
Lemma 2: Given xn + yn = zn 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 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 proof.]
(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(10) So, we can conclude that k is an n-power. [By the lemma above]
(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.
(11) Which means that there exists u such that un = k.
(12) And we get Dn * un = Zn
(13) And (Du)n = Zn
(14) Implying that Du = Z and multiplying by c that du=z.
(15) Which proves that d divides z.