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 * 7^2ev² + u)(s² + 3 *7^2ev - u) = 64 * 7^4e-1v⁴
(b) gcd(s² + 3 * 7^2ev² + u , s² + 3 *7^2ev - u) = 2ⁿ where n ≥ 0.
(c) v is odd
Then there exists a,b such that:
(a) (8s + 32b² - 3 *7^2ea²)(8s - 32b² + 3* 7^2ea²) = 7^4e-1a⁴
(b) a,b are odd
(c) gcd(a,b)=1

Proof:

(1) From (b), we know that 7^4e-1 divides either (s² + 3 * 7^2ev² + u) or (s² + 3 *7^2ev - 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 * 7^2ea²b² ± u = 7^4e-1Aa⁴
(b) s² + 3 * 7^2ea²b² ± 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 * 7^2ea²b² = 7^4e-1Aa⁴ + Bb⁴

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

(8) From #7, we get:
2s² = -6 * 7^2ea²b² + 7^4e-1Aa⁴ + Bb⁴

(9) So that:
s² = -3 * 7^2ea²b² + 7^4e-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(7^2e)a²b² + 7^4e-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 * 7^2ea²b² + 7^4e-1(A/2)a⁴ + (B/2)b⁴ =
= -3 * 7^2ea²b² + 7^4e-1(2/2)a⁴ + (32/2)b⁴ =
= -3 * 7^2ea²b² + 7^4e-1a⁴ + 16b4.

(15) This can be adjusted to:
s² + 3 * 7^2ea²b² = 7^4e-1a⁴ + 16b⁴

(16) Multiplying both sides by 64 gives us:
64² + 6 * 7^2e*32a²b² = 7^4e-1*64a⁴ + 64*16b⁴

(17) Subtracting 64*16b⁴ from both sides gives us:
64s² + 6 * 7^2e * 32a²b² - 64 * 16b⁴ = 7^4e-1*64a⁴

(18) Since (32b² - 3 * 7^2ea²)² = 64*16b⁴ - 6*7^2e32a²b² + 3²*7^4ea⁴, we can change (#17) into:

64s² - (32b² - 3*7^2ea²)² =
= 7^4e-1 * 64a⁴ -3²7^4ea⁴ = (64 - 9*7)7^4e-1a⁴ = 7^4e-1a⁴

(19) But then we have:

(8s + (32b² - 3*7^2ea²)(8s - 32b² - 3*7^2ea²) =
= (8s + 32b² - 3*7^2ea²)(8s - 32b² + 3*7^2ea2) = 7^4e-1a⁴

QED

Lemma 4: if (8s + 32b² - 3 *7^2ea²)(8s - 32b² + 3* 7^2ea²) = 7^4e-1a⁴
with a,b are odd, gcd(a,b)=1, and 7 doesn't divide s then gcd(8s + 32b² - 3 *7^2ea², 8s - 32b² + 3* 7^2ea²) = 1.

(1) Assume that the gcd(8s + 32b² - 3 *7^2ea², 8s - 32b² + 3* 7^2ea²) 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*7^2ea² =
= (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*7^2ea² =
= (even)(odd) ± (even)(odd)² ±3*7^2ea² =
= (even) ± (even) ± (odd) = odd.

(5) This means that p must divide s.

(6) By our original assumption p must divide 7^4e-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*7^2ea²

(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 * 7^2ev² + u)(s² + 3 *7^2ev - u) = 64 * 7^4e-1v⁴
(b) gcd(s² + 3 * 7^2ev² + u , s² + 3 *7^2ev - 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 *7^2ea²)(8s - 32b² + 3* 7^2ea²) = 7^4e-1a⁴
a,b are odd
gcd(a,b)=1

(3) From Lemma 4, we know that:
gcd(8s + 32b² - 3 *7^2ea², 8s - 32b² + 3* 7^2ea²) = 1.

(4) From this, we know that 7^4e-1 divides either 8s + 32b² - 3 *7^2ea² or 8s - 32b² + 3* 7^2ea².

(5) Also, we know that there must exist two odd integers c,d such that:
(a) a = cd
(b) 7^4e-1c⁴ divides either 8s + 32b² - 3 *7^2ea² or 8s - 32b² + 3* 7^2ea² 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*7^2ea² ≡ (0) ± (0) & #177; 3 * 7^2ea² ≡ ±3 * 7^2ea² (mod 8)

(8) From Lemma 1, since 7^2e = (7^e)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