## Friday, May 27, 2005

### Fermat's Last Theorem: n = 3: Step 4

Today's blog continues the proof for Fermat's Last Theorem n=3 which was started in a previous blog.

Lemma: Given the following conditions:

(a) x,y,z is a solution to x3 + y3 = z3
(b) two values of x,y,z are derived from p+q,p-q
(c) p,q coprime
(d) p,q opposite parity
(e) p,q positive
(f) 2p*(p2 + 3q2) is a cube.
(g) gcd(2p,p2 + 3q2) = 3

Then:
There exists a smaller solution A,B,C such that A3 + B3 = C3.

(1) First, we note that 3 divides p but not q. Since 3 divides 2p [see #g] and p,q are coprime [see #c].

(2) So, there exists s such that:
p = 3s and
2p * (p2 + 3q2) = 2p*(3s*3s + 3q2) = 2*3s*(3*3s2 + 3q2) =
= 32*2s(3s2 + q2)

(3) Now 32*2s, 3s2 + q2 are relatively prime since:
(a) 3 doesn't divide q [step #1]

(b) so 3 doesn't divide 3s2 + q2

(c) Since p=3s [#2], s is the same parity as p which means that s,q have opposite parity [see #d in given].

(d) But if s,q have opposite parity, then 2 doesn't divide 3s2 + q2 since it must be odd.

(e) Finally, gcd(s,q)=1 since gcd(p,q)=1. [see #c in the given]

(4) So, 32*2s, 3s2 + q2 are cubes [By Relatively Prime Divisor Lemma] since 32*2s(3s2 + q2) = 2p * (p2 + 3q2) [see #2] and 2p * (p2 + 3q2) is a cube [see #f in given].

(5) Then, there exists a,b such that:

q = a3 - 9ab2
s = 3a2b - 3b3
gcd(a,b)=1

since:

gcd(q,s)=1 [See #3e above]
q,s have opposite parities [see #3c above]
q2 + 3s2 is a cube [see #4 above]

[See here for the lemma that establishes this]

(6) From, this we can show that 2b, a-b, a+b are cubes
(a) a,b have different parity since q,s have different parity [see #3c above] since:

Case I: a,b are both even

q = a3 - 9ab2 = even - even = even.
s = 3a2b - 3b3 = even - even = even

which is impossible since q,s have different parity.

Case II: a,b are both odd

q = a3 - 9ab2 = odd - odd = even
s = 3a2b - 3b3 = odd - odd = even.

which is impossible since q,s have different parity.

(b) a + b, a - b are odd since a,b have different parity so 2 is coprime.

(c) b is coprime to a + b and a - b, otherwise, it would divide a which goes against gcd(a,b)=1

(d) a + b, a - b are coprime since any common factor would be odd and divide both a and b since 2a = a + b + a - b and 2b = a + b - (a - b)

(e) 32*2s is a cube [see #4 above] so 32*2s =32*2[3a2b - 3b3] = 33*2[a2b - b3] = 33(2b)(a+b)(a - b) is a cube.

(f) But if 33(2b)(a+b)(a - b) is a cube, then (2b)(a+b)(a - b) is a cube.

(g) But if (2b)(a+b)(a - b) is a cube and gcd(2b,a+b,a-b)=1 [by #6b,#6c,#6d], then by the Relatively Prime Divisor Lemma, 2b, a+b, and a-b are all cubes.
(7) But this means there exists A,B,C such that:
A3 = 2b
B3 = a - b
C3 = a + b

(8) Which means that there is another solutions to Fermat's Last Theorem n=3 since:

A3 = 2b = a + b - (a - b) = C3 - B3

(9) Now, since:

C3 = a + b which is less than s = (3b)(a - b)(a + b) which is less than p = 3s which is less than either x3 or z3 since either z3 = (2p)*(p2 + 3q2) or x3 = (2p) * (p2 + 3q2).

(10) So, a solution here leads necessarily to a smaller solution of FLT n=3.

QED Anonymous said...

Hi Larry,

What should equate to (2b)(a+b)(a-b) in step VI? s=(3b)(a+b)(a-b).

Thanks.
-Jay

Larry Freeman said...

Hi Jay,

Thanks very much for your question. I looked over the proof and decided that the clarity of its logic could be greatly improved.