## Thursday, January 19, 2006

### 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 Sn = Sn-1p - Sn-2q + Sn-3r is a recurrence relationship where Sn = 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 Sn recurrence formula works for all values between 1 and n.

(4) Sn+1 = Snp - Sn-1q + Sn-2r =
(xn + yn + zn)(x + y + z) - (xn-1 + yn-1 + zn-1)(xy + xz + yz) + (xn-2 + yn-2 + zn-2)(xyz)

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

(xn-1 + yn-1 + zn-1)(xy + xz + yz) =
= xny + xnz + xn-1yz + xyn + xyn-1z + ynz + xyzn-1 + xzn + yzn

(xn-2 + yn-2 + zn-2)(xyz) =
= xn-1yz + xyn-1z + xyzn-1

Combining all this together gives us:
Sn+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

#### 1 comment:

Scouse Rob said...

In step (2) should:

(x^2 + y^2 + z^2)(xy + xz + yz)
= x^3y + x^3z + x^2yz + xy^2 + xy^2z + y^3z + xyz^2 + xz^3 + yz^3

be

(x^2 + y^2 + z^2)(xy + xz + yz)
= x^3y + x^3z + x^2yz + xy^3 + xy^2z + y^3z + xyz^2 + xz^3 + yz^3

Rob

Rob