## Monday, January 23, 2006

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

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 :
(a) 2 * 74e-1 *A4= s2 + 3 * 72ev2 ± u
(b) 2 * B4
= s2 + 3 * 72ev2 ± u
(c) s, A, B, v are an integers
(d) gcd(A,B) = 1
(e) v is even
(f) s is odd
Then
B is not even

Proof:

(1) Assume B is even

(2) Then A is odd

(3) Adding the two values together gives us:
( s2 + 3 * 72ev2 ± u) + ( s2 + 3 * 72ev2 ± u) = 2s2 + 2*3*72ev2 = 2 * 74e-1 *A4 + 2 * B4

(4) Dividing both sides by 2 gives us:
s2 + 3*72ev2 = 74e-1 * A4 + B4

(5) Subtracting 3*72ev2 from both sides gives us:
s2 = -3*72ev2 + 74e-1*A4 + B4

(6) But from #5 and from a previous result (see here), we get:
s2 ≡ (-3)(1)(v2) + (-1)4e-1(1) + (B2)2 (mod 8)

(7) Also, from a previous result (see here), we know that B2 ≡ 0 or 4 (mod 8).

(8) So that (B2)2 ≡ 0 (mod 8) since:

(a) (0)2 ≡ 0 (mod 8)

(b) (4)2 ≡ 16 ≡ 0 (mod 8)

(9) So applying (#9) to (#7) gives us:
s2 ≡ (-3)v2 + (-1)4e-1(1) + (0) ≡ (-3)v2 - 1.

(10) But since v is even, there exists v' such that v = 2v' and this means that:
v2 ≡ (2v')2 ≡ 4v'2 ≡ 0 since:

(a) So, v'2 ≡ 0 or 4 (mod 8)

(b) (4)(0) ≡ 0 (mod 8)

(c) (4)(4) ≡ 16 ≡ 0 (mod 8)

(11) Combining (#11) with (#10) gives us that:
s2 ≡ -1 (mod 8).

(12) But this is impossible since s2 ≡ 1 (mod 8) by a previous result (see here)

(13) So we have a contradiction and we reject our assumption.

QED

Lemma 2: if:
(a) 2 * 74e-1 *A4= s2 + 3 * 72ev2 ± u
(b) 2 * B4
= s2 + 3 * 72ev2 ± u
(c) s, A, B, v are an integers
(d) gcd(A,B) = 1
(e) v is even
(f) s is odd
(g) A is even
Then
There exists an integer A'' such that A = 8A''

Proof:

(1) B is odd [since A is even and gcd(A,B)=1]

(2) Since A is even, there exists a value A' such that A= 2A'

(3) Since s is odd and from a previous result (see here), we have:
s2 ≡ 1 (mod 8)

(4) Adding (a) and (b), we get:
2*s2 = -3 *2* 72ev2 + 2*74e-1A4 + 2*B4

(5) Since 2v=AB, we can also write:
s2 = -3 * 72e([1/2]AB)2 + 74e-1A4 + B4 =
= -3*72e(A'B)2 + 74e-1A4 + B4

(6) From (#5) and a previous result (see here):
s2 ≡ (-3)(1)(A')2(1) + (-1)4e-1(0) + (1) ≡
≡ (-3)(A')2 + 1 (mod 8)

(7) Combining (#6) with (#3) gives us:
s2 ≡ (-3)(A') + 1 ≡ 1 (mod 8)

(8) This implies:
(-3)(A')2 ≡ 0 (mod 8)

(9) Which mean that A' must be divisible by 4 since:

(-3)(0)2 ≡ 0 (mod 8)
(-3)(1)2 ≡ -3 (mod 8) which doesn't work.
(-3)(2)2 ≡ (-3)(4) ≡ -4 (mod 8) which doesn't work
(-3)(3)2 ≡ (-3)(9) ≡ -27 ≡ -3 (mod 8) which doesn't work
(-3)(4)2 ≡ (-3)(16) ≡ -48 ≡ 0 (mod 8)
(-3)(5)2 ≡ (-3)(25) ≡ -75 ≡ -3 (mod 8) which doesn't work
(-3)(6)2 ≡ (-3)(36) ≡ -108 ≡ -4 (mod 8) which doesn't work
(-3)(7)2 ≡ (-3)(49) ≡ -147 ≡ -3 (mod 8) which doesn't work

(10) So there exists A'' such that A' = 4A'' = 8A

QED

Lemma 3: if :
(a) 2 * 74e-1 *A4= s2 + 3 * 72ev2 ± u
(b) 2 * B4
= s2 + 3 * 72ev2 ± u
(c) s, A, B, v are an integers
(d) gcd(A,B) = 1
(e) v is even
(f) s is odd
(g) A is even
Then
There exists A'' such that A=8A'' and:
(s - B2 + 3*8*72eA''2)(s + B2 - 3*8*72eA''2) = 74e-1*4*(2A'')4

Proof:

(1) From Lemma 2 above, we know that there exists A'' such that A=8A''.

(2) Adding (a) and (b) together gives us:
s2 = -3 * 72e([1/2]AB)2 + 74e-1A4 + B4 =
= -3*72e(A'B)2 + 74e-1A4 + B4

(3) -3*16*72e(A'')2B2 + 74e-1*84*(A'')4 + B4

(4) Now,
(B2 - 3*8*72e(A'')2)2 = B4 - 2*3*8*72e(A'')2B2 + 9*64*74e(A'')4

(5) So,
s2 - (B2 - 3*8*72e(A'')2)2 = 74e-1*84*(A'')4 - 32*82*74e(A'')4 =
= 74e-1(84*(A'')4 - 32*82*7(A'')4) =
= 74e-1*82(82(A'')4 - 32*7(A'')2) =
= 74e-1*82(64(A'')4 - 63(A'')4) =
= 74e-1*82(A'')4

(6) Putting it altogether gives us:
(S - B2 + 3*8*72e(A'')2)(S + B2 + 3*8*72e(A'')2) = 74e-1*82*(A'')4 =
= 74e-1*4*(2A'')4

QED

Lemma 4: x2 ≡ 0, 1, 2, or 4 (mod 7)

Proof:

(0)2 ≡ 0 (mod 7)
(1)2 ≡ 1 (mod 7)
(2)2 ≡ 4 (mod 7)
(3)2 ≡ 9 ≡ 2 (mod 7)
(4)2 ≡ 16 ≡ 2 (mod 7)
(5)2 ≡ 25 ≡ 4 (mod 7)
(6)2 ≡ 36 ≡ 1 (mod 7)

QED

Lemma 5: if :
(a) 2 * 74e-1 *A4= s2 + 3 * 72ev2 ± u
(b) 2 * B4
= s2 + 3 * 72ev2 ± u
(c) s, A, B, v are an integers
(d) gcd(A,B) = 1
(e) v is even
(f) s is odd
(g) A is even
Then
There exists A'' such that A=8A'' and:
(s - B2 + 3*8*72eA''2)(s + B2 - 3*8*72eA''2) ≠ 74e-1*4*(2A'')4

Proof:

(1) Assume that (s - B2 + 3*8*72eA''2)(s + B2 - 3*8*72eA''2) = 74e-1*4*(2A'')4.

(2) Both (s - B2 + 3*8*72eA''2) and (s + B2 + 3*8*72eA''2) are even since:

(a) s is odd, B is odd (since A is even and gcd(A,B)=1.)

(b) odd - odd + even = even

(c) odd + odd + even = even

(3) gcd(s - B2 + 3*8*72eA''2 , s + B2 - 3*8*72eA''2) = 2 since:

(a) Let p be a prime that divides both and which does not equal 2.

(b) Since p divides both it also divides (s - B2 + 3*8*72eA''2 ) + (s + B2 + 3*8*72eA''2) = 2s.

(c) Since p is odd, it divides s.

(d) By assumption p must divide 74e-1*4*(2A'')4.

(e) But this means p must divide 74e-1 or A''.

(f) p ≠ 7 since it divides s and s is not divisible by 7.

(g) p cannot divide A'' since by S - B2 + 3*8*72eA''2, it would have to divide B but this is impossible since gcd(A,B)=1.

(h) So we have a contradiction and decide that p=2.

(i) Let f be the gcd. By (h), it must be a power of 2.

(j) By the logic in (b), it must divide 2s which means that it must = 2 since s is odd.

(4) From (#1), we can conclude that:

(a) 74e-1 divides 1 of the two multiples.

(b) We know that there exists c,d such that cd=2A'' and c4 divides 1 of the multiplies and d4 divides the other and gcd(c4,d4) = 1.

(c) So, we can assume that:
s ± B2 ± 3*8*72eA''2 = 2c4

s ± B2 ± 3*8*72eA''2 = 74e-12d4

(5) Subtracting these two equations so that we remove the s gives us:
± 2B2 = ±6*8*72eA''2 - 2c4 + 74e-12d4

(6) Dividing both sides by 2 gives us:
± B2 = ± 6*4*72eA''2 - c4 + 74e-1d4 =
= ± 6*72e(2A'')2 - c4 + 74e-1d4 =
= ± 6*72ec2d2 - c4 + 74e-1d4

(7) We know that 7 does not divide B

If 7 divided B, then 7 would divide c from (#6), and then it would divide s from (#4c) which is impossible since s is not divisible by 7.

(8) And we also know that 7 does not divide c.

If 7 divided c, then 7 would divide B from (#6) and the same argument applies.

(9) So ± B2 + c4 ≡ 0 (mod 7) from (#6)

(10) But then we have ± B2 + (1, 2, or 4) which means ± must actually be "-".

(11) This then gives us: -B2 = -6*72ec2d2 -c4 + 74e-1d4

(12) Multiplying -1 to each side gives us:
B2 = c2 + 6*72ec2d2 - 74e-1d4

(13) But this equation leads to an infinite descent by a previous lemma (see step #5 in Lemma 1 here) since:

v = AB/2 = 4A''B ≥ 2cd which is greater than d.

(14) So we reject our assumption at (1).

QED

Lemma 6: 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
(d) t = 7ev
(e) gcd(t,s)=1
Then
v is not even.

Proof:

(1) Assume that v is even.

(2) Then t must be even and s must be odd from (d) and (e).

(3) u must also be odd since:

(a) Assume u is even.

(b) (s2 + 3 * 72ev2 + u)(s2 + 3 *72ev - u) = 64 * 74e-1v4

(c) (odd2 + odd*odd*even + even)(odd2 + odd*odd*even - even) =
(odd + even + even)(odd + even - even) = (odd)(odd) = odd

(d) Which contradicts (b) so we reject our assumption (a).

(4) Since v is even, we know that there exists v' such that:
v = 2v'

(5) Since u is odd, we know that both (s2 + 3 * 72ev2 + u) and (s2 + 3 *72ev - u) are even since:

odd*odd + odd*odd*even*even + odd = odd + even + odd = even.

(6) This gives us the following:
(s2 + 3 * 72ev2 + u)(s2 + 3 *72ev - u) = (2)(2) * 74e-1(16)v4

(7) From (b) above we know that 74e-1 divides either one or the other since if it did not divide only 1, there would a common factor of 7 which is not the case.

(8) From (#6) and (#7), we can posit that there exists A,B such that:
AB = 2v

And (2v)4 = A4B4

(9) And putting this into (6) gives us:
(s2 + 3 * 72ev2 + u)(s2 + 3 *72ev - u) = (2)(2) * 74e-1(16)v4 = (2)(2)* 74e-1A4B4

Where:

2*74e-1A4 divides either (s2 + 3 * 72ev2 + u) or (s2 + 3 *72ev - u) and
2*B4 divides the other one.

(10) We note further that gcd( s2 + 3 * 72ev2 + u, s2 + 3 *72ev - u) can be at most 2 since:

(a) Let f = the greatest common denominator.

(b) f must divide (s2 + 3 * 72ev2 + u) - (s2 + 3 *72ev - u) = 2u

(c) We know from our assumption that f must be a power of 2.

(d) We also know that u is odd.

(e) This means that f must = 2.

(11) So, we have the following:

(a) s2 + 3 * 72ev2 ± u = 2 * 74e-1A4

(b) s2 + 3 * 72ev2 ± u = 2 * B4

(c) we can further conclude that gcd(A,B) = 1 since:

If an odd prime divides gcd(A,B), then this same odd prime would divide (s2 + 3 * 72ev2 + u) and (s2 + 3 *72ev - u) which goes against our assumption.

If 2 divides gcd(A,B), then then we would have 4 would divide both but this impossible since the gcd can be at most 2.

(d) There are two possible cases to consider:

(i) A is odd, B is even
(ii) A is even, B is odd

We know that they cannot both be even from (#11c). We know that one must be even since they are equal to 2v (#8)

(12) B is not even (see Lemma 1 above)

(13) So A is even.

(14) So (s - B2 + 3*8*72eA''2)(s + B2 - 3*8*72eA''2) = 74e-1*4*(2A'')4 [From Lemma 3 above]

(15) But simultaneously, from Lemma 5 above we have:
(s - B2 + 3*8*72eA''2)(s + B2 - 3*8*72eA''2) ≠ 74e-1*4*(2A'')4

(16) So we have a contradiction and we reject (#1).

QED