## Thursday, January 19, 2006

### Fermat's Last Theorem: Proof for n=7: Existence M,Q

In today's blog, I will continue the proof for Fermat's Last Theorem n=7.

The details for today are based on Paulo Ribenboim's Fermat's Last Theorem for Amateurs.

Lemma 1: If x,y,z are integers such that:
(a) p = x + y + z ≠ 0
(b) q = xy + xz + yz
(c) r = xyz
Then:
x7 + y7 + z7 = p7 - 7(pq - r)(p4 - p2q + q2) + 7(pq - r)2p.

Proof:

(1) From Newton's Formula for sums, we get
x7 + y7 + z7 = p7 - 7p5q + 7p4r + 14p3q2 - 21p2qr - 7pq3 + 7pr2 + 7q2r.

(2) We know that:
(-7)(pq - r)(p4 - p2q + q2) = -7p5q + 7p3q2 - 7pq3 + 7p4r - 7p2qr + 7q2r.

(3) From the Binomial Theorem, we know that:
(pq - r)2p = 7p3q2 - 14p2qr + 7pr2

(4) Combining #1, #2, and #3 gives us:
p7 - 7(pq -r)(p4 - p2q + q2) + 7(pq -r)2p =
p7 - 7p5q + 7p4r + 14p3q2 - 21p2qr - 7pq3 + 7pr2 + 7q2r.

QED

Lemma 2: If x,y,z are integers such that:
(a) x7 + y7 + z7 = 0
(b) xyz ≠ 0
(c) x + y + z ≠ 0
Then there exist rational numbers M,Q such that:
(a) p = x + y + z
(b) q = xy + xz + yz
(c) r = xyz
(d) m = pq - r

(e) Q = q/p2
(f) M = m/p3
(g) M2 - M(1 - Q + Q2) + 1/7 = 0

Proof:

(1) Let p = x + y + z.

(2) Let q = xy + xz + yz.

(3) Let r = xyz

(4) From Lemma 1 above, we know that:
x7 + y7 + z7 = p7 - 7(pq - r)(p4 - p2q + q2) + 7(pq - r)2p.

(5) Let m = pq - r

(6) #4 gives us:
x7 + y7 + z7 = p7 - 7m(p4 - p2q + q2) + 7m2p.

(7) Let Q = q/p2 so that q = p2Q

(8) Let M = m/p3 so that m = p3M

(9) Then, #6 gives us:
x7 + y7 + z7 = p7-7p3M(p4 - p4Q + p4Q2) + 7p7M2 = p7 - 7p7M - 7p7MQ -7p7MQ2 + 7p7M2

(10) Dividing both sides by 7p7 (since x7 + y7 + z7 = 0) gives us:

1/7 -M -MQ - MQ2 + M2 = M2 - M(1 - Q + Q2) + 1/7 = 0

QED

Scouse Rob said...

In Lemma 1 step (3) should:

(pq - r)^2p = 7p^3q^2 - 14p^2qr + 7pr^2

be

7(pq - r)^2p = 7p^3q^2 - 14p^2qr + 7pr^2

Rob

Scouse Rob said...

In Lemma 2 step (9) should

p^7-7p^3M(p^4 - p^4Q + p^4Q^2) + 7p^7M^2 = p^7 - 7p^7M - 7p^7MQ -7p^7MQ^2 + 7p^7M^2

be

p^7-7p^3M(p^4 - p^4Q + p^4Q^2) + 7p^7M^2 = p^7 - 7p^7M + 7p^7MQ -7p^7MQ^2 + 7p^7M^2

Rob

Scouse Rob said...

In Lemma 2 step (10) should:

1/7 -M - MQ - MQ^2 + M^2 = M^2 - M(1 - Q + Q^2) + 1/7 = 0

be

1/7 - M + MQ - MQ^2 + M^2 = M^2 - M(1 - Q + Q^2) + 1/7 = 0

Rob