Proof of Fermat’s Last Theorem and Beal’s Conjecture
A.O. Awojobi
Fermat’s Last Theorem
Pn + Qn ≠ Rn, n > 2 if P, Q, R and n are all positive integers.
Andrew Wiles produced a lengthy proof of over 100 pages in the mid
1990s.
Beal’s Conjecture
If Ax + By = Cz, x > 2, y > 2 and z > 2 where A, B, C, x, y and z are
positive integers then A, B and C must have a highest common factor
> 1.
The left hand side of the equation can be rewritten as Ax + By
= (Ax/n)n + (By/n)n = [Ax/n + By/n][(Ax/n)n-1 - (Ax/n)n-2 By/n +
(Ax/n)n-3 (By/n)2 -... - (Ax/n)2 (By/n)n-3 + Ax/n(By/n)n-2 - (By/n)n-1] where
n > 2 is an odd integer. Beal’s conjecture can be rewritten as
Cz - By = Ax and the left hand side can further be rewritten as
Cz - By = (Cz/n)n - (By/n)n = [Cz/n - By/n][(Cz/n)n-1 + (Cz/n)n-2 By/n +
(Cz/n)n-3 (By/n)2 +... + (Cz/n)2 (By/n)n-3 + Cz/n(By/n)n-2 + (By/n)n-1] where
n > 2 is an integer.
Let it be initially assumed that A and B have a highest common factor
= 1 and Ax + By = (Ax/n)n + (By/n)n = [Ax/n + By/n][(Ax/n)n-1 (Ax/n)n-2 By/n + (Ax/n)n-3 (By/n)2 -...- (Ax/n)2 (By/n)n-3 + Ax/n(By/n)n-2 (By/n)n-1] = Mn. Let M = d - e, Ax/n = hd, By/n = ie where M is a
positive integer, d and e are positive rational numbers. Therefore
[Ax/n + By/n][(Ax/n)n-1 - (Ax/n)n-2 By/n + (Ax/n)n-3(By/n)2 -...
- (Ax/n)2 (By/n)n-3 + Ax/n(By/n)n-2 - (By/n)n-1]/M = Mn-1 i.e.
[(hd + ie)/(d - e)]hn-1dn-1 - [(hd + ie)/(d - e)]hn-2 i dn-2 e +
[(hd + ie)/(d - e)]hn-3 i2 dn-3 e2 - … - [(hd + ie)/(d - e)]h2 in-3 d2 en-3 +
[(hd + ie)/(d - e)]h in-2 d en-2- [(hd + ie)/(d - e)] in-1 en-1
= dn-1 - (n-1)dn-2 e + (𝑛 −2 1) dn-3 e2 - … - (𝑛 −2 1) d2 en-3 + (n-1) d e n-2 - en-1
If this is an identity then corresponding coefficients on both sides can
be made equal to each other. Equating the first and last coefficients on
the left hand side to the first and last binomial coefficient of 1 on the
right hand side implies that h = i. Equating the second and second to
Apr 30 2016 13:56:31 EDT
Version 1 - Submitted to PROC
the last coefficients on the left hand side to the second and second to
the last binomial coefficient of n-1 on the right hand side implies that
h = i once again. If this is carried on h = i every time because of the
symmetry of the binomial coefficients. If h = i this implies that all
coefficients on the left hand side are the same. If corresponding
coefficients on both sides are equated given that h = i then conflicting
values of h = i begin to arise. This is therefore not an identity and the
left hand side cannot ever be equal to the right hand side. Therefore
Ax + By ≠ Mn if A and B have a highest common factor = 1.
Let it be initially assumed that C and B have a highest common factor
= 1 and Cz - By = (Cz/n)n - (By/n)n = [Cz/n - By/n][(Cz/n)n-1 + (Cz/n)n-2 By/n
+ (Cz/n)n-3 (By/n)2 +... + (Cz/n)2 (By/n)n-3 + Cz/n(By/n)n-2 + (By/n)n-1] = Ln.
Let L = f + g, By/n = jf and Cz/n = kg where L is a positive integer,
f and g are positive rational numbers. Therefore [Cz/n - By/n][(Cz/n)n-1 +
(Cz/n)n-2 By/n + (Cz/n)n-3 (By/n)2 +... + (Cz/n)2 (By/n)n-3 + Cz/n(By/n)n-2 +
(By/n)n-1]/L = Ln-1 i.e. [(kg – jf)/(f + g)]kn-1 gn-1 +
[(kg – jf)/(f + g)]kn-2 j gn-2 f + [(kg – jf)/(f + g)]kn-3 j2 gn-3 f2 + … +
[(kg – jf)/(f + g)]k2 jn-3 g2 fn-3 + [(kg – jf)/(f + g)]k jn-2 g fn-2 + [(kg –
jf)/(f + g)] jn-1 fn-1 = gn-1 + (n-1) gn-2 f + (𝑛 −2 1) gn-3 f2 + … + (𝑛 −2 1) g2 fn-3
+ (n-1) g f n-2 + fn-1
If this is an identity then corresponding coefficients on both sides can
be made equal to each other. Equating the first and last coefficients on
the left hand side to the first and last binomial coefficient of 1 on the
right hand side implies that j = k. Equating the second and second to
the last coefficients on the left hand side to the second and second to
the last binomial coefficient of n-1 on the right hand side implies that
j = k once again. If this is carried on j = k every time because of the
symmetry of the binomial coefficients. If j = k this implies that all
coefficients on the left hand side are the same. If corresponding
coefficients on both sides are equated given that j = k then conflicting
values of j = k begin to arise. This is therefore not an identity and the
left hand side cannot ever be equal to the right hand side. Therefore
Cz - By ≠ Ln if C and B have a highest common factor = 1.
Apr 30 2016 13:56:31 EDT
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Proof of Fermat’s Last Theorem
It has just been shown that (Cz/n)n - (By/n)n ≠ Ln. Let Cz/n and By/n be
integers. Let tCz/n = R, tBy/n = Q and tL = P, where t is the highest
common factor of P, Q and R. When each term in the inequality
(Cz/n)n - (By/n)n ≠ Ln is multiplied by tn then Rn - Qn ≠ Pn i.e. Fermat’s
Last theorem is proved. For the special case when n = 2 i.e. the
Pythagorean theorem, there will be no conflicting values of j = k. j = k
will have a particular value since there are only 2 terms on each side
of what is therefore an identity.
Proof of Beal’s Conjecture
It has been shown previously that Ax + By = (Ax/n)n + (By/n)n ≠ Mn,
n>2 is an odd integer and Cz - By = (Cz/n)n - (By/n)n ≠ Ln if A and B
have a highest common factor = 1, C and B have a highest common
factor = 1 respectively. This essentially proves that Beal’s conjecture
is true. It should now be clear why all examples of Beal’s conjecture
that have x, y and z to be different all originate from Beal conjecture
examples where 2 of x, y or z are the same e.g. 76 + 77 = 983 can be
rewritten as 493 + 77 = 983. 24210 + 24211 = 1756925 can be rewritten
as 585645 + 24211 = 1756925 or 2425 + 2426 = 7265 if each of the terms
is divided by 2425.
Apr 30 2016 13:56:31 EDT
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