Playing with Fermat's Last Theorem

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Some results from playing with Fermat's Last Theorem  
 
Some results from playing with Fermat's Last Theorem  
by Will Johnson, wjhonson@aol.com copyright 2001-4,
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by Will Johnson, [mailto:wjhonson@aol.com wjhonson@aol.com] copyright 2001-4, all rights reserved.
all rights reserved.
+
  
 
(Condition 1) x^n + y^n = z^n
 
(Condition 1) x^n + y^n = z^n
(Condition 2) n prime and not 2
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(Condition 3) x,y,z natural numbers, > 0
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(Condition 2) n is prime and is not 2
 +
 
 +
(Condition 3) x,y,z are natural numbers and are > 0
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(Condition 4) gcd(x,y) = gcd(x,z) = gcd(y,z) = 1 that is, they share no common factors.
 
(Condition 4) gcd(x,y) = gcd(x,z) = gcd(y,z) = 1 that is, they share no common factors.
  
Theorem:  Given the above conditions, show that z-x and z-y
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Theorem:  Given the above conditions, show that z-x and z-y are each and independently, either nth powers or n times a nth power.
are both and independently, either nth powers or
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n times a nth power.
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From Condition 3 and Condition 2
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From Condition 3 and Condition 2 we have that
 
x^n, y^n and z^n are all greater than zero.
 
x^n, y^n and z^n are all greater than zero.
  
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Expanding (d+x)^n and cancelling the last term, we get
 
Expanding (d+x)^n and cancelling the last term, we get
(Equation 5) y^n = d^n + nd^(n-1)x + (n 2)d^(n-2)x^2 + ...
+
(Equation 5) y^n = d^n + nd^(n-1)x + (n 2)d^(n-2)x^2 + ... + (n 2) (d^2)x^(n-2) + ndx^(n-1)
+ (n 2) (d^2)x^(n-2) + ndx^(n-1)
+
  
Since d divides every term on the right, it must also evenly
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Since d divides every term on the right, it must also evenly divide what's on the left therefore  
divide what's on the left therefore  
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(Result 2) d divides y^n
 
(Result 2) d divides y^n
  
 
Let (Equation 6) d = p(1)^l(1) * p(2)^l(2) * ... * p(k)^l(k)
 
Let (Equation 6) d = p(1)^l(1) * p(2)^l(2) * ... * p(k)^l(k)
be the unique prime factorization of d
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be the unique prime factorization of d
 +
 
 
Then since gcd(d,x) = 1 this implies that
 
Then since gcd(d,x) = 1 this implies that
 
(Result 3) gcd(p(j),x) = 1 for all p(j) in d
 
(Result 3) gcd(p(j),x) = 1 for all p(j) in d
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Let (Equation 7) y(1)^t(1) * y(2)^t(2) * ... * y(m)^t(m)
 
Let (Equation 7) y(1)^t(1) * y(2)^t(2) * ... * y(m)^t(m)
be the unique prime factorization of y. Then  
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be the unique prime factorization of y. Then  
(Result 5) each p(j) = y(o) for some o and  
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(Result 5) each p(j) = y(o) for some o and therefore p(j) has a relationship to t(o)*n
therefore p(j) has a relationship to t(o)*n
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If l(j) < t(o)n this implies p(j) divides nx^(n-1) which implies
 
If l(j) < t(o)n this implies p(j) divides nx^(n-1) which implies
 
(Result 6) p(j) = n or p(j) divides x^(n-1)
 
(Result 6) p(j) = n or p(j) divides x^(n-1)
 +
 
But, by Result 3, p(j) cannot divide x^(n-1) since to do so, since P(j) is prime, it must divide x which it is forbidden to do and therefore p(j) = n
 
But, by Result 3, p(j) cannot divide x^(n-1) since to do so, since P(j) is prime, it must divide x which it is forbidden to do and therefore p(j) = n
  
If l(j) > t(o)n this implies p(j) divides some other factor of y
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If l(j) > t(o)n this implies p(j) divides some other factor of y as well which contradicts the Prime Factorization Theorem, since each y(j) was chosen to be independent of any other.
as well which contradicts the Prime Factorization Theorem, since each y(j) was chosen to be independent of any other.
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The third case is left:
 
The third case is left:
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Now if p(j) = n then n divides d and also n divides y which means that  
 
Now if p(j) = n then n divides d and also n divides y which means that  
n^n divides y^n
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n^n divides y^n therefore the right hand side of Equation 5 must also be divisible by n^n.  Trying each successive n^q one by one we see that actually n^(n-1) must divide d since none of these n's can divide x, since n divides d and result 1 says that x and d have no common factors, and exactly one can divide the n coefficient in the last term of Equation 5.
therefore the right hand side of Equation 5 must also be  
+
 
divisible by n^n.  Trying each successive n^q one by one we  
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(Result 8) Therefore, we have shown, for each element of the prime factorization of d, that either:
see that actually n^(n-1) must divide d since none of these
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p(j) = n AND l(j) = t(o)n - 1
n's can divide x, since n divides d and result 1 says that x and d have no common factors, and exactly one can divide the n coefficient in the last term of Equation 5.
+
  
(Result 8) Therefore, we have shown, for each element of the
 
prime factorization of d, that either:
 
p(j) = n AND l(j) = t(o)n - 1
 
 
OR l(j) = t(o)n
 
OR l(j) = t(o)n
  
Now this is true for every element of the prime factorization
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Now this is true for every element of the prime factorization of d, therefore:
of d, therefore:
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(Result 9)  if n divides d the other elements of its prime factorization must be nth powers so there exists an e such that d = n^(tn-1) * e^n
(Result 9)  if n divides d the other elements of its prime factorization must be nth
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powers so there exists an e such that d = n^(tn-1) * e^n
+
  
 
Now tn-1 = n*(t-1) + (n-1) so this can become
 
Now tn-1 = n*(t-1) + (n-1) so this can become
d = n^(n-1) * (en^(t-1))^n = n^(n-1) * E
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d = n^(n-1) * (en^(t-1))^n = n^(n-1) * E
if n does not divide d all elements of d's prime factorization must be nth powers so
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if n does not divide d all elements of d's prime factorization must be nth powers so there exists an e such that d = e^n
there exists an e such that d = e^n
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Now we arbitrarily chose y to act upon, so therefore choosing
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Now we arbitrarily chose y to act upon, so therefore choosing x has the same result.
x has the same result.
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Reducing equation 5, mod n, we get y^n congruent to d^n mod n
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Reducing equation 5, mod n, we get y^n is congruent to d^n mod n
 
(Result 10) n divides d if and only if n divides y
 
(Result 10) n divides d if and only if n divides y
  
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The case n divides y is analogous to the case n divides x
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The case n divides y is analogous to the case n divides x so we don't have to consider them both.
so we don't have to consider them both.
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Case 1 n divides y
 
Case 1 n divides y
 
(Equation 1) x^n + y^n = (d+x)^n-x^n =  (n^(tn-1)*e^n + x)^n = z
 
(Equation 1) x^n + y^n = (d+x)^n-x^n =  (n^(tn-1)*e^n + x)^n = z
  Also y = n^t * Y for some Y and t > 0
+
 
x^n + (n^t * Y)^n = (n^(tn-1)*e^n + x)^n
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Also y = n^t * Y for some Y and t > 0
(n^t * Y)^n = ....n^(tn)e^n * x
+
x^n + (n^t * Y)^n = (n^(tn-1)*e^n + x)^n
Y^n = ....e^n * x
+
(n^t * Y)^n = ....n^(tn)e^n * x
 +
Y^n = ....e^n * x
 
(Equation 2) x^n + y^n = (d+y)^n - y^n = (f^n + y)^n = z
 
(Equation 2) x^n + y^n = (d+y)^n - y^n = (f^n + y)^n = z
  
 
Setting these equal we have  (n^(tn-1)*e^n + x)^n = (f^n + y)^n
 
Setting these equal we have  (n^(tn-1)*e^n + x)^n = (f^n + y)^n
Taking the nth root we get
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Taking the nth root we get n^(tn-1)*e^n + x = f^n + y = z
n^(tn-1)*e^n + x = f^n + y = z
+
  
 
Now since n^t divides y, let y = Yn^t then substituting we have
 
Now since n^t divides y, let y = Yn^t then substituting we have
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Let n = 2 to get
 
Let n = 2 to get
 
2^(2t-1)*e^2 + x = f^2 + 2^tY
 
2^(2t-1)*e^2 + x = f^2 + 2^tY
for the case z-x = n we have n^(tn-1)*e^n = n implies tn-1 = 1
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for the case z-x = n we have n^(tn-1)*e^n = n implies tn-1 = 1 which is only solved if t = 1 and n = 2
which is only solved if t = 1 and n = 2
+
 
 
let t = 1 and n = 3 then 3^2 * e^3 = 9e^3
 
let t = 1 and n = 3 then 3^2 * e^3 = 9e^3
 
x^3 + y^3 = (9e^3 + x)^3 = 9^3e^9 + 3*9^2e^6x + 3*9e^3x^2 + x^3
 
x^3 + y^3 = (9e^3 + x)^3 = 9^3e^9 + 3*9^2e^6x + 3*9e^3x^2 + x^3

Revision as of 16:37, 19 September 2008

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