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