## Sunday, January 22, 2006

### Fermat's Last Theorem: Proof for n=7: v cannot be odd

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 is odd, then x2 ≡ 1 (mod 8).

(1) Since x is odd, there exists a value x' such that x = 2x' + 1.

(2) x2 = (2x' + 1)2 = 4x'2 + 4x' + 1 = 4(x'2 + x') + 1

(3) Now x'2 + x' is even since:

Case I: x' is odd

odd + odd2 = odd + odd = even.

Case II: x' is even

even + even2 = even + even = even

(4) So, we can suppose that there exists x'' such that x'2 + x' = 2x''

(5) Which gives us 4(2x'') + 1 = 8x'' + 1

QED

Lemma 2: if x is even, then x2 ≡ 0 or 4 (mod 8)

(1) From modular arithmetic, we only need to show this is true where x = 0, 2, 4, or 6 since any other even number will equate to these results. (See here for review of modular arithmetic)

(2) Now, we try out the 4 possible even values:

(0)2 ≡ 0 (mod 8)
(2)2 ≡ 4 (mod 8)
(4)2 ≡ 16 ≡ 0 (mod 8)
(6)2 ≡ 36 ≡ 4 (mod 8)

QED

Lemma 3: if:
(a) (s2 + 3 * 72ev2 + u)(s2 + 3 *72ev - u) = 64 * 74e-1v4
(b) gcd(s2 + 3 * 72ev2 + u , s2 + 3 *72ev - u) = 2n where n ≥ 0.
(c) v is odd
Then there exists a,b such that:
(a) (8s + 32b2 - 3 *72ea2)(8s - 32b2 + 3* 72ea2) = 74e-1a4
(b) a,b are odd
(c) gcd(a,b)=1

Proof:

(1) From (b), we know that 74e-1 divides either (s2 + 3 * 72ev2 + u) or (s2 + 3 *72ev - u) [Otherwise, 7 could be a common factor which is not the case.

(2) We can suppose two values A,B such that AB=64 and A divides the same value as (1) while B divides the other one.

(3) Also, we can suppose two values a,b such that v = ab where a divides the same value as (1) while b divides the other.

(4) We know that a,b are odd since v is odd and odd = odd * odd.

(5) We also know that gcd(a,b)=1 from (b). If this wasn't the case, then there would be an odd prime that was a common factor which (b) tells us is not the case.

(6) Putting this altogether, we get:
(a) s2 + 3 * 72ea2b2 ± u = 74e-1Aa4
(b) s2 + 3 * 72ea2b2 ± u = Bb4

NOTE: I used ± since we don't which one is (a) and which one is (b). The important point is that they complementary so if (a) were "+", then b would be "-".

(7) Adding (a) and (b) together gives us:
2s2 + 6 * 72ea2b2 = 74e-1Aa4 + Bb4

NOTE: Because (a) and (b) are complementary, the u cancels out.

(8) From #7, we get:
2s2 = -6 * 72ea2b2 + 74e-1Aa4 + Bb4

(9) So that:
s2 = -3 * 72ea2b2 + 74e-1(A/2)a4 + (B/2)b4

(10) Since s is an integer, we know that A/2 and B/2 must be integers.

(11) So we can deduce from (9)

s2 ≡ -3(72e)a2b2 + 74e-1(A/2)a4 + (B/2)b4 (mod 8)

(12) Applying Lemma 1 and the fact that a,b are odd gives us:

s2 ≡ (-3)(1)(1)(1) + (-1)4e-1(A/2)(1) + (B/2)(1) ≡

≡ (-3) + (-1)(A/2) + (B/2) ≡ -3 - (A/2) + (B/2) (mod 8)

(13) Now from (#12), (#2), and (#10) we can deduce that A = 2, B = 32 since:

The possible values for A,B are (2,32), (4,16), (8, 8), (16,4), (32,2)

If S is odd, then S2 ≡ 1 (mod 8) from Lemma 1.
If S is even, then S2 ≡ 0 or 4(mod 8) from Lemma 2.

So, now we show that of the 5 possible values for A,B only (2,32) works:

Case I: A=2, B = 32 is possible since:

-3 -(A/2) + (B/2) ≡ -3 -(2/2) + (32/2) ≡ -3 - 1 + 16 ≡ 20 ≡ 4 (mod 8)

Case II: A=4, B=16 is impossible since:

-3-(A/2) + (B/2) ≡ -3 -(4/2) + (16/2) ≡ -3 -2 + 8 ≡ 3 (mod 8)

Case III: A=8, B=8 is impossible since:

-3-(A/2) + (B/2) ≡ -3 - (8/2) + (8/2) ≡ -3 -4 + 4 ≡ -3 (mod 8)

Case IV: A=16, B = 4 is impossible since:

-3-(16/2) + (4/2) ≡ -3 -8 + 2 ≡ -9 ≡ -1 (mod 8)

Case V: A=32, B = 2 is impossible since:

-3 - (32/2) + (2/2) ≡ -3 -16 + 1 ≡ -18 ≡ -2 (mod 8)

(14) So using (#13) with (#19), we get:

s2 = -3 * 72ea2b2 + 74e-1(A/2)a4 + (B/2)b4 =
= -3 * 72ea2b2 + 74e-1(2/2)a4 + (32/2)b4 =
= -3 * 72ea2b2 + 74e-1a4 + 16b4.

(15) This can be adjusted to:
s2 + 3 * 72ea2b2 = 74e-1a4 + 16b4

(16) Multiplying both sides by 64 gives us:
642 + 6 * 72e*32a2b2 = 74e-1*64a4 + 64*16b4

(17) Subtracting 64*16b4 from both sides gives us:
64s2 + 6 * 72e * 32a2b2 - 64 * 16b4 = 74e-1*64a4

(18) Since (32b2 - 3 * 72ea2)2 = 64*16b4 - 6*72e32a2b2 + 32*74ea4, we can change (#17) into:

64s2 - (32b2 - 3*72ea2)2 =
= 74e-1 * 64a4 -3274ea4 = (64 - 9*7)74e-1a4 = 74e-1a4

(19) But then we have:

(8s + (32b2 - 3*72ea2)(8s - 32b2 - 3*72ea2) =
= (8s + 32b2 - 3*72ea2)(8s - 32b2 + 3*72ea2) = 74e-1a4

QED

Lemma 4: if (8s + 32b2 - 3 *72ea2)(8s - 32b2 + 3* 72ea2) = 74e-1a4
with a,b are odd, gcd(a,b)=1, and 7 doesn't divide s then gcd(8s + 32b2 - 3 *72ea2, 8s - 32b2 + 3* 72ea2) = 1.

(1) Assume that the gcd(8s + 32b2 - 3 *72ea2, 8s - 32b2 + 3* 72ea2) is greater than 1.

(2) Then there exists a prime p that divides both.

(3) Adding the two values together, we get that p divides 16s which means that p divides 16 or p divides s by Euclid's Lemma.

(4) First, we note that p ≠ 2 since:

(a) s is either even or odd.

(b) if s is even, then 8s ± 32b2 ±3*72ea2 =
= (even)(even) ± (even)(odd)2 ±(odd)*(odd)2e(odd)2 =
= (even) ± (even) ± (odd) = odd.

(c) Likewise, if s is odd, then then 8s ± 32b2 ±3*72ea2 =
= (even)(odd) ± (even)(odd)2 ±3*72ea2 =
= (even) ± (even) ± (odd) = odd.

(5) This means that p must divide s.

(6) By our original assumption p must divide 74e-1a4.

(7) But since p divides s, we know that p ≠ 7 since 7 doesn't divide s.

(8) This means that p must divide a again by Euclid's Lemma.

(7) But combining (#5) and (#2) gives us that p must divide 32b2 - 3*72ea2

(8) Which would mean since p divides a and p ≠ 2, then p must divide b2 which means that p divides b.

(9) But this is impossible since gcd(a,b) = 1.

QED

Lemma 5: If there exists s,u,v such that:
(a) (s2 + 3 * 72ev2 + u)(s2 + 3 *72ev - u) = 64 * 74e-1v4
(b) gcd(s2 + 3 * 72ev2 + u , s2 + 3 *72ev - u) = 2n where n ≥ 0.
(c) 7 doesn't divide s
Then
v is not odd.

Proof:

(1) Assume that v is odd.

(2) From Lemma 3, we know that there exists a,b such that:
(8s + 32b2 - 3 *72ea2)(8s - 32b2 + 3* 72ea2) = 74e-1a4
a,b are odd
gcd(a,b)=1

(3) From Lemma 4, we know that:
gcd(8s + 32b2 - 3 *72ea2, 8s - 32b2 + 3* 72ea2) = 1.

(4) From this, we know that 74e-1 divides either 8s + 32b2 - 3 *72ea2 or 8s - 32b2 + 3* 72ea2.

(5) Also, we know that there must exist two odd integers c,d such that:
(a) a = cd
(b) 74e-1c4 divides either 8s + 32b2 - 3 *72ea2 or 8s - 32b2 + 3* 72ea2 while d4 divides the other one.

(6) From #3, we know that gcd(c,d)=1 [otherwise we violate step (#3).]

(7) Adding the two values in (#4) together gives us:
d4 ≡ 8s ± 32b2 ± 3*72ea2 ≡ (0) ± (0) & #177; 3 * 72ea2 ≡ ±3 * 72ea2 (mod 8)

(8) From Lemma 1, since 72e = (7e)2 and since a is odd, this gives us:
d4 ≡ ±(3)(1)(1) ≡ ± 3 (mod 8)

(9) But this is impossible since d is odd, d2 is odd and (odd)2 ≡ 1 (mod 8) from Lemma 1.

(10) Therefore we have a contradiction and we reject our assumption.

QED

Scouse Rob said...

In Lemma 3 step (14) should

(14) So using (#13) with (#19), we get:

be

(14) So using (#13) with (#9), we get:

Rob

Scouse Rob said...

In Lemma 3 step (16) should:

64^2 + 6 * 72^e*32a^2b^2 = 7^4e-1*64a^4 + 64*16b^4

be

64^s2 + 6 * 72^e*32a^2b^2 = 7^4e-1*64a^4 + 64*16b^4

Scouse Rob said...

In Lemma 3 step (19) should:

(8s + (32b^2 - 3*7^2ea^2)(8s - 32b^2 - 3*7^2ea^2)

be

(8s + (32b^2 - 3*7^2ea^2)(8s - (32b^2 - 3*7^2ea^2))