Fermat's Last Theorem: Proof for n=7: Newton's Formula

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.

Today's proof uses a method that was first used by Sir Isaac Newton.

Lemma 1: Newton's Method:
(a) Let p = x + y + z
(b) Let q = xy + xz + yz
(c) Let r = xyz
(d) Let S₁ = p
(e) Let S₂ = p² - 2q
(f) Let S₃ = p³ - 3pq +3r
Then
For n ≥ 4 S_n = S_n-1p - S_n-2q + S_n-3r is a recurrence relationship where S_n = xⁿ + yⁿ + zⁿ.

Proof:

(1) We start by noting that this result works for S₁, S₂, and S₃:

S₁ = p = x + y + z = x¹ + y¹ + z¹

S₂ = p² - 2q = (x + y + z)² - 2(xy + xz + yz) =
x² + xy + xz + y² + xy + yz + z² + xz + yz - 2xy -2xz - 2yz = x² + y² + z²

S₃ = p³ - 3pq + 3r = (x + y + z)³ - 3(x + y + z)(xy + xz + yz) + 3(xyz) = x³ + y³ + x³

(2) Let's expand S₄:

S₄ = S₃p - S₂q + S₁r =
= (x³ + y³ + z³)(x + y + z) - (x² + y² + z²)(xy + yz + xz) + (x + y + z)(xyz)

(x³ + y³ + z³)(x + y + z) =
= x⁴ + y⁴ + z⁴ + x³y + x³z + xy³ + y³z + xz³ + yz³

(x² + y² + z²)(xy + xz + yz) =
= x³y + x³z + x²yz + xy² + xy²z + y³z + xyz² + xz³ + yz³

(x + y + z)(xyz) = x²yz + xy²z +xyz²

Combining all the above gives us:

S₄ = x⁴ + y⁴ + z⁴

(3) Since it's true for S₄, we know that there exists some n ≥ 4 such that the S_n recurrence formula works for all values between 1 and n.

(4) S_n+1 = S_np - S_n-1q + S_n-2r =
(xⁿ + yⁿ + zⁿ)(x + y + z) - (x^n-1 + y^n-1 + z^n-1)(xy + xz + yz) + (x^n-2 + y^n-2 + z^n-2)(xyz)

(xⁿ + yⁿ + zⁿ)(x + y + z) =
= xⁿ⁺¹ + yⁿ⁺¹ + zⁿ⁺¹ + xⁿy + xⁿz + xyⁿ + yⁿz + xzⁿ + yzⁿ

(x^n-1 + y^n-1 + z^n-1)(xy + xz + yz) =
= xⁿy + xⁿz + x^n-1yz + xyⁿ + xy^n-1z + yⁿz + xyz^n-1 + xzⁿ + yzⁿ

(x^n-2 + y^n-2 + z^n-2)(xyz) =
= x^n-1yz + xy^n-1z + xyz^n-1

Combining all this together gives us:
S_n+1 = xⁿ⁺¹ + yⁿ⁺¹ + zⁿ⁺¹

(5) By the Principle of Induction, we are done.

QED

Lemma 2: If there exists integers x,y,z such that:
(a) x⁷ + y⁷ + z⁷ = 0
(b) x + y + z ≠ 0
(c) xyz ≠ 0
(d) p = x + y + z
(e) q = xy + xz + yz
(f) r = xyz
Then
x⁷ + y⁷ + z⁷ = p⁷ - 7p⁵q + 7p⁴r + 14p³q² - 21p²qr - 7pq³ + 7pr² + 7q²r.

Proof:

(1) By Lemma 1 above, we get:
S₇ = S₆p - S₅q + S₄r = x⁷ + y⁷ + z⁷

S₆ = p⁶ - 6p⁴q + 6p³r + 9p²q² - 12pqr - 2q³ + 3r²

S₅ = p⁵ -5p³q + 5p²r + 5pq² - 5qr

S₄ = p⁴ - 4p²q + 4pr + 2q²

(2) Combining these all together gives us:

x⁷ + y⁷ + z⁷ = p⁷ - 7p⁵q + 7p⁴r + 14p³q² - 21p²qr - 7pq³ + 7pr² + 7q²r.

QED