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*Post by Simon Roberts*Yes finally: THIRD PAGE INDUCED (PLUS THE ASIDE) (ROUGH PROOF COMPLETE)

https://www.dropbox.com/sh/602u4zmdr1b4goz/AACKp6RaA3KD7H6i4Ci15QxPa?dl=0

Thanks

Simon

*Post by Simon Roberts*Hello and a plea,

Can these be completed (if correct) ; Fermat's?

If so then_please you are welcome to finish and post if willing.

https://www.dropbox.com/sh/602u4zmdr1b4goz/AACKp6RaA3KD7H6i4Ci15QxPa?dl=0

Thank You,

Simon C. Roberts

ASIDE OR APPENDIX

Version 3.1

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p_j = 1 (mod q) where,

a^q - b^q = (a - b)[Product_j (p_j)],

GCD(a,b) = 1,

q is an odd prime and,

p_j are odd primes.

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Proof.

(0.)

(a^q - b^q)/(a - b) is a positive odd integer.

(I leave this conclusion for the reader).

(1.)

(a^q - b^q)/(a - b) =

a^(q-1) + ba^(q-2) + b^2a^(q-3) + ... +

a^2b^(q-3) + ab^(q-2) + b^(q-1).

(2.)

[(a^q - b^q)/(a - b)] - qa^(q-1) =

a^(q-1) - a^(q-1) + (b-a)a^(q-2) + (b^2-a^2)a^(q-3) + ... +

a^2(b^(q-3)-a^(q-3)) + a(b^(q-2)-a^(q-2) + (b^(q-1)-a^(q-1)).

(a-b) divides the RHS of (2.).

Therefore,

(3.)

(a-b) | [(a^q - b^q)/(a - b) - qa^(q-1)].

But,

(4.)

GCD( (a^q - b^q)/(a - b), qa^(q-1) ) = 1.

Therefore,

(5.)

GCD((a^q - b^q)/(a - b), (a-b)) = 1.

(6.)

(a^q - b^q) = (a-b) * PRODUCT_j[p_j] =

PRODUCT_i[r_i] * PRODUCT_j[p_j] where,

a - b = PRODUCT_i[r_i], r_i are odd primes and,

(a^q - b^q)/(a - b) = PRODUCT_j[p_j], p_j are odd primes.

(7.)

For any j and for any i, p_j =/= r_i. This follows from

(5.)

GCD((a^q - b^q)/(a - b), (a-b)) = 1.

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Inductive Proof.

(8.0)

a =/= b (mod p_i) for all p_i.

(8.1)

a^q = b^q (mod p_i) for any p_i.

(8.2.1) from (8.0 and 8.1)

a^(q+1) =/= b^(q+1) (mod p_i) for all p_i.

(8.2.2) from (8.0 and 8.1)

a^(q-1) =/= b^(q-1) (mod p_i) for all p_i.

(8.3)

a^2 =/= b^2 (mod p_i) for all p_i.

This follows from (8.2.1) and/or (8.2.2).

That is, 2 divides either or both (q-1) or (q+1).

(8.3.1)

a^(q-2) =/= b^(q-2) (mod p_i) for all p_i

(this follows from (8.3) and (8.1)).

(8.3.2)

a^(q+2) =/= b^(q+2) (mod p_i) for all p_i

(this also follows from (8.3) and (8.1)).

Now,

3 divides [(q-2) or (q-1) or (q+1) or (q+2)].

(3 does not divide q unless q =3. Then proof would be near complete).

Therefore,

(8.3.3)

a^3 =/= b^3 (mod p_i) for all p_i.

................................................

Induction.

Assume,

(8.m)

a^m =/= b^m (mod p_i) for all p_i

and m < q.

Also it is also assumed, at this point,

a^k =/= b^k (mod p_i) for k: 1 =< k =< m < (q-1).

(8.m.1)

a^(q-k) =/= b^(q-k) (mod p_i) for all p_i

(this follows from (8.m) and (8.1)).

(8.m.2)

a^(q+k) =/= b^(q+k) (mod p_i) for all p_i

(this follows from (8.m) and (8.1)).

(8.m.3)

[(m+1)|(q+k)] OR [(m+1)|(q-k)] for all k: 1 =< k =< m < (q-1);

(q-k) to (q+k) has 2m + 1 values in succession from (q-m) to (q+m),

including q.

(m+1) must divide one of the values in this range.

Therefore,

(8.[m+1])

a^(m+1) =/= b(m+1) (mod p_i) for all p_i.

End of induction.

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(9.0)

a^n =/= b^n (mod p_i) for all p_i

n: 1 =< n < q,

Also,

(10.0)

a^n =/= b^n (mod p_i) for all p_i

n: n =/= wq, w is any positive integer.

Recall,

(6.)

(a^q - b^q) = PRODUCT_i[r_i] * PRODUCT_j[p_j].

a - b = PRODUCT_i[r_i].

Now,

(11.)

a^(p_j-1) = b^(p_j-1) (mod p_j) for all p_j,

(this follows from Fermat's Little Theorem).

(12.)

p_j - 1 =/= n;

if p_j - 1 = n, then

a^(n) = b^(n) (mod p_j) for all p_j. Contradiction.

n is equal to any positive integer other than wq.

w is any positive integer.

This implies,

p_j - 1 = wq

and therefore,

(13.)

p_j = 1 (mod q) for all p_j.

(6.1)

(a^q - b^q)/(a-b) = PRODUCT_j[p_j]

End Proof.

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-Simon Roberts

***@gmail.com