The details for today are based on Paulo Ribenboim's Fermat's Last Theorem for Amateurs.

Lemma 1: if :

(a) 2 * 7

^{4e-1}*A

^{4}= s

^{2}+ 3 * 7

^{2e}v

^{2}± u

(b) 2 * B

^{4}= s

^{2}+ 3 * 7

^{2e}v

^{2}± 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:

( s

^{2}+ 3 * 7

^{2e}v

^{2}± u) + ( s

^{2}+ 3 * 7

^{2e}v

^{2}± u) = 2s

^{2}+ 2*3*7

^{2e}v

^{2}= 2 * 7

^{4e-1}*A

^{4}+ 2 * B

^{4}

(4) Dividing both sides by 2 gives us:

s

^{2}+ 3*7

^{2e}v

^{2}= 7

^{4e-1}* A

^{4}+ B

^{4}

(5) Subtracting 3*7

^{2e}v

^{2}from both sides gives us:

s

^{2}= -3*7

^{2e}v

^{2}+ 7

^{4e-1}*A

^{4}+ B

^{4}

(6) But from #5 and from a previous result (see here), we get:

s

^{2}≡ (-3)(1)(v

^{2}) + (-1)

^{4e-1}(1) + (B

^{2})

^{2}(mod 8)

(7) Also, from a previous result (see here), we know that B

^{2}≡ 0 or 4 (mod 8).

(8) So that (B

^{2})

^{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:

s

^{2}≡ (-3)v

^{2}+ (-1)

^{4e-1}(1) + (0) ≡ (-3)v

^{2}- 1.

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

v

^{2}≡ (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:

s

^{2}≡ -1 (mod 8).

(12) But this is impossible since s

^{2}≡ 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 * 7

^{4e-1}*A

^{4}= s

^{2}+ 3 * 7

^{2e}v

^{2}± u

(b) 2 * B

^{4}= s

^{2}+ 3 * 7

^{2e}v

^{2}± 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:

s

^{2}≡ 1 (mod 8)

(4) Adding (a) and (b), we get:

2*s

^{2}= -3 *2* 7

^{2e}v

^{2}+ 2*7

^{4e-1}A

^{4}+ 2*B

^{4}

(5) Since 2v=AB, we can also write:

s

^{2}= -3 * 7

^{2e}([1/2]AB)

^{2}+ 7

^{4e-1}A

^{4}+ B

^{4}=

= -3*7

^{2e}(A'B)

^{2}+ 7

^{4e-1}A

^{4}+ B

^{4}

(6) From (#5) and a previous result (see here):

s

^{2}≡ (-3)(1)(A')

^{2}(1) + (-1)

^{4e-1}(0) + (1) ≡

≡ (-3)(A')

^{2}+ 1 (mod 8)

(7) Combining (#6) with (#3) gives us:

s

^{2}≡ (-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 * 7

^{4e-1}*A

^{4}= s

^{2}+ 3 * 7

^{2e}v

^{2}± u

(b) 2 * B

^{4}= s

^{2}+ 3 * 7

^{2e}v

^{2}± 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 - B

^{2}+ 3*8*7

^{2e}A''

^{2})(s + B

^{2}- 3*8*7

^{2e}A''

^{2}) = 7

^{4e-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:

s

^{2}= -3 * 7

^{2e}([1/2]AB)

^{2}+ 7

^{4e-1}A

^{4}+ B

^{4}=

= -3*7

^{2e}(A'B)

^{2}+ 7

^{4e-1}A

^{4}+ B

^{4}

(3) -3*16*7

^{2e}(A'')

^{2}B

^{2}+ 7

^{4e-1}*8

^{4}*(A'')

^{4}+ B

^{4}

(4) Now,

(B

^{2}- 3*8*7

^{2e}(A'')

^{2})

^{2}= B

^{4}- 2*3*8*7

^{2e}(A'')

^{2}B

^{2}+ 9*64*7

^{4e}(A'')

^{4}

(5) So,

s

^{2}- (B

^{2}- 3*8*7

^{2e}(A'')

^{2})

^{2}= 7

^{4e-1}*8

^{4}*(A'')

^{4}- 3

^{2}*8

^{2}*7

^{4e}(A'')

^{4}=

= 7

^{4e-1}(8

^{4}*(A'')

^{4}- 3

^{2}*8

^{2}*7(A'')

^{4}) =

= 7

^{4e-1}*8

^{2}(8

^{2}(A'')

^{4}- 3

^{2}*7(A'')

^{2}) =

= 7

^{4e-1}*8

^{2}(64(A'')

^{4}- 63(A'')

^{4}) =

= 7

^{4e-1}*8

^{2}(A'')

^{4}

(6) Putting it altogether gives us:

(S - B

^{2}+ 3*8*7

^{2e}(A'')

^{2})(S + B

^{2}+ 3*8*7

^{2e}(A'')

^{2}) = 7

^{4e-1}*8

^{2}*(A'')

^{4}=

= 7

^{4e-1}*4*(2A'')

^{4}

QED

Lemma 4: x

^{2}≡ 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 * 7

^{4e-1}*A

^{4}= s

^{2}+ 3 * 7

^{2e}v

^{2}± u

(b) 2 * B

^{4}= s

^{2}+ 3 * 7

^{2e}v

^{2}± 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 - B

^{2}+ 3*8*7

^{2e}A''

^{2})(s + B

^{2}- 3*8*7

^{2e}A''

^{2}) ≠ 7

^{4e-1}*4*(2A'')

^{4}

Proof:

(1) Assume that (s - B

^{2}+ 3*8*7

^{2e}A''

^{2})(s + B

^{2}- 3*8*7

^{2e}A''

^{2}) = 7

^{4e-1}*4*(2A'')

^{4}.

(2) Both (s - B

^{2}+ 3*8*7

^{2e}A''

^{2}) and (s + B

^{2}+ 3*8*7

^{2e}A''

^{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 - B

^{2}+ 3*8*7

^{2e}A''

^{2}, s + B

^{2}- 3*8*7

^{2e}A''

^{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 - B

^{2}+ 3*8*7

^{2e}A''

^{2}) + (s + B

^{2}+ 3*8*7

^{2e}A''

^{2}) = 2s.

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

(d) By assumption p must divide 7

^{4e-1}*4*(2A'')

^{4}.

(e) But this means p must divide 7

^{4e-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 - B

^{2}+ 3*8*7

^{2e}A''

^{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) 7

^{4e-1 }divides 1 of the two multiples.

(b) We know that there exists c,d such that cd=2A'' and c

^{4}divides 1 of the multiplies and d

^{4}divides the other and gcd(c

^{4},d

^{4}) = 1.

(c) So, we can assume that:

s ± B

^{2}± 3*8*7

^{2e}A''

^{2 }= 2c

^{4}

s ± B

^{2}± 3*8*7

^{2e}A''

^{2}= 7

^{4e-1}2d

^{4}

(5) Subtracting these two equations so that we remove the s gives us:

± 2B

^{2}= ±6*8*7

^{2e}A''

^{2}- 2c

^{4}+ 7

^{4e-1}2d

^{4}

(6) Dividing both sides by 2 gives us:

± B

^{2}= ± 6*4*7

^{2e}A''

^{2}- c

^{4}+ 7

^{4e-1}d

^{4}=

= ± 6*7

^{2e}(2A'')

^{2}- c

^{4}+ 7

^{4e-1}d

^{4}=

= ± 6*7

^{2e}c

^{2}d

^{2}- c

^{4}+ 7

^{4e-1}d

^{4}

(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 ± B

^{2}+ c

^{4}≡ 0 (mod 7) from (#6)

(10) But then we have ± B

^{2}+ (1, 2, or 4) which means ± must actually be "-".

(11) This then gives us: -B

^{2}= -6*7

^{2e}c

^{2}d

^{2}-c

^{4}+ 7

^{4e-1}d

^{4}

(12) Multiplying -1 to each side gives us:

B

^{2}= c

^{2}+ 6*7

^{2e}c

^{2}d

^{2}- 7

^{4e-1}d

^{4}

(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) (s

^{2}+ 3 * 7

^{2e}v

^{2}+ u)(s

^{2}+ 3 *7

^{2e}v - u) = 64 * 7

^{4e-1}v

^{4}

(b) gcd(s

^{2}+ 3 * 7

^{2e}v

^{2}+ u , s

^{2}+ 3 *7

^{2e}v - u) = 2

^{n}where n ≥ 0.

(c) 7 doesn't divide s

(d) t = 7

^{e}v

(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) (s

^{2}+ 3 * 7

^{2e}v

^{2}+ u)(s

^{2}+ 3 *7

^{2e}v - u) = 64 * 7

^{4e-1}v

^{4}

(c) (odd

^{2}+ odd*odd*even + even)(odd

^{2}+ 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 (s

^{2}+ 3 * 7

^{2e}v

^{2}+ u) and (s

^{2}+ 3 *7

^{2e}v - u) are even since:

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

(6) This gives us the following:

(s

^{2}+ 3 * 7

^{2e}v

^{2}+ u)(s

^{2}+ 3 *7

^{2e}v - u) = (2)(2) * 7

^{4e-1}(16)v

^{4}

(7) From (b) above we know that 7

^{4e-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}= A

^{4}B

^{4}

(9) And putting this into (6) gives us:

(s

^{2}+ 3 * 7

^{2e}v

^{2}+ u)(s

^{2}+ 3 *7

^{2e}v - u) = (2)(2) * 7

^{4e-1}(16)v

^{4}= (2)(2)* 7

^{4e-1}A

^{4}B

^{4}

Where:

2*7

^{4e-1}A

^{4}divides either (s

^{2}+ 3 * 7

^{2e}v

^{2}+ u) or (s

^{2}+ 3 *7

^{2e}v - u) and

2*B

^{4}divides the other one.

(10) We note further that gcd( s

^{2}+ 3 * 7

^{2e}v

^{2}+ u, s

^{2}+ 3 *7

^{2e}v - u) can be at most 2 since:

(a) Let f = the greatest common denominator.

(b) f must divide (s

^{2}+ 3 * 7

^{2e}v

^{2}+ u) - (s

^{2}+ 3 *7

^{2e}v - 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) s

^{2}+ 3 * 7

^{2e}v

^{2}± u = 2 * 7

^{4e-1}A

^{4}

(b) s

^{2}+ 3 * 7

^{2e}v

^{2}± u = 2 * B

^{4}

(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 (s

^{2}+ 3 * 7

^{2e}v

^{2}+ u) and (s

^{2}+ 3 *7

^{2e}v - 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 - B

^{2}+ 3*8*7

^{2e}A''

^{2})(s + B

^{2}- 3*8*7

^{2e}A''

^{2}) = 7

^{4e-1}*4*(2A'')

^{4}[From Lemma 3 above]

(15) But simultaneously, from Lemma 5 above we have:

(s - B

^{2}+ 3*8*7

^{2e}A''

^{2})(s + B

^{2}- 3*8*7

^{2e}A''

^{2}) ≠ 7

^{4e-1}*4*(2A'')

^{4}

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

QED

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