Fermat's Last Theorem: n = 3: Step 1

http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-n-3-step-1.html

Fermat's Last Theorem: n = 3: Step 1

This blog will explain the proof for the first step in Fermat's Last Theorem: n = 3. For the full proof, please start here. The details for this proof come from Leonhard Euler who was the first to make progress on Fermat's Last Theorem.

The first step will be stated as a lemma:

Lemma: x³ + y³ = z³ has a solution only if there exists p,q such that:

(a) gcd(p,q)=1
(b) p,q are positive
(c) p,q have opposite parity (that is, one is even, one is odd)
(d) 2p*(p² + 3q²) is a cube.

(1) We can assume that x,y,z are coprime.

(2) Since, x,y,z are coprime, we know that at most one of them is even.

(3) We also know that at least one of them is even since if x and y are odd, then z must be even.

(4) So, we now divide this proof up into Case I: z is even and Case II: x is even. Since x,yare symmetrical, Case II will also cover the case where y is even.

Case I: z is even

(1) Then x,y are odd.

(2) x + y and x - y are even.

(3) Let 2p = x + y and 2q = x - y

(4) Then:
x = (1/2)[x + y + ( x - y )] = p + q
y = (1/2)[x + y - ( x - y )] = p - q

(5) Now gcd(p,q) = 1

(a) Assume f = gcd(p,q) is greater than 1
(b) Then there exists P,Q such that p = fP, q = fQ
(c) But then f divides both x,y since x = f(P + Q), y = f(P - Q)
(d) This is a contradiction since x,y are coprime.

(6) We can assume that p,q are positive

(a) From before p = (1/2)(x + y), q = (1/2)(x - y)
(b) x cannot equal y since they are coprime.
(c) if x + y is negative, then substitute (-x),(-y) since x³ + y³ = z³ implies that(-x)³ + (-y)³ = (-z)³
(d) if y is greater than x then flip x,y since they are symmetric.
(e) This covers all cases.

(7) p,q have to have different parities since x,y are both odd.

(8) Finally, 2p*(p² + 3q²) is a cube.

z³ = x³ + y³ =
= (x + y)(x² - xy + y²) =
= (p + q + p - q)[(p+q)² - (p+q)(p-q) + (p-q)²] =
= 2p(p² + 3q²)

Case II: x is even

(1) Then z,y are odd since they are coprime to x.

(2) z+y, z-y are both even.

(3) There exists p,q such that 2p = (z - y), 2q = (z + y)

(4) And z = (1/2)[(z - y) + (z + y)] = (1/2)(2p + 2q) = p + q, y = (1/2)[(z + y) - (z - y)] = (1/2)(2q - 2p) = q-p

(5) p,q have opposite parity since z,y are odd.

(6) gcd(p,q)= 1 [Same argument as Case I]

(7) p,q are positive [Same argument as Case I]

(8) 2p*(p² + 3q²) is a cube.

x³ = z³ - y³ =
= (z - y)(z² + zy + y²) =
= [q + p - (q - p)][(q+p)² + (q+p)(q-p) + (q-p)²] =
= 2p*(p² + 3q²)

QED