Playing with Fermat's Last Theorem
Some results from playing with Fermat's Last Theorem by Will Johnson, firstname.lastname@example.org copyright 2001-4, all rights reserved.
(Condition 1) x^n + y^n = z^n
(Condition 2) n is prime and is not 2
(Condition 3) x,y,z are natural numbers and are > 0
(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 are each and independently, either nth powers or n times a nth power.
From Condition 3 and Condition 2 we have that x^n, y^n and z^n are all greater than zero.
From Condition 1 subtract x^n to get
- (Equation 2) y^n = z^n - x^n
Since z must be greater than either y or x, there exists a d, natural number, such that
- (Equation 3) z = d+x
If gcd(d,x) > 1 then, d and x have a common factor, but since z = d + x this would imply that z also shares this same common factor and from Condition 4 we have that gcd(x,z) = 1 which violates this assumption and therefore it must be that
- (Result 1) gcd(d,x) = 1
Substituting Equation 3 into Equation 2 we get
- (Equation 4) y^n = (d+x)^n - x^n
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 + ... + (n 2) (d^2)x^(n-2) + ndx^(n-1)
Since d divides every term on the right, it must also evenly divide what's on the left therefore
- (Result 2) d divides y^n
Let (Equation 6) d = p(1)^l(1) * p(2)^l(2) * ... * p(k)^l(k) be the unique prime factorization of d
Then since gcd(d,x) = 1 this implies that
- (Result 3) gcd(p(j),x) = 1 for all p(j) in d
Now each p(j)^l(j) divides y^n therefore, since each p(j) is prime,
- (Result 4) each p(j) divides y
Let (Equation 7) y(1)^t(1) * y(2)^t(2) * ... * y(m)^t(m) be the unique prime factorization of y. Then
- (Result 5) each p(j) = y(o) for some o and therefore p(j) has a relationship to t(o)*n
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)
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 as well which contradicts the Prime Factorization Theorem, since each y(j) was chosen to be independent of any other.
The third case is left:
- (Result 7) l(j) = t(o)n
Now if p(j) = n then n divides d and also n divides y which means that 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.
(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
Now this is true for every element of the prime factorization of d, therefore:
- (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
Now tn-1 = n*(t-1) + (n-1) so this can become
- 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 there exists an e such that d = e^n
Now we arbitrarily chose y to act upon, so therefore choosing x has the same result.
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
FINAL RESULT Therefore (Result 11):
- if n^t divides y then z = n^(tn-1) * e^n + x for some e ;
- if n^t does NOT divide y then z = e^n + x for some e.
In other words, z - x is either a perfect nth power or the n-1th power of n times a perfect nth power.
Another way to state this is:
- x^n + y^n = (x + t^n/n)^n if n divides y
- x^n + y^n = (x + t^n)^n if n does not divide y
The case n divides y is analogous to the case n divides x so we don't have to consider them both.
Case 1 n divides y (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 (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
Setting these equal we have (n^(tn-1)*e^n + x)^n = (f^n + y)^n Taking the nth root we get n^(tn-1)*e^n + x = f^n + y = z
Now since n^t divides y, let y = Yn^t then substituting we have n^(tn-1)*e^n + x = f^n + n^tY
Let n = 2 to get 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 which is only solved if t = 1 and n = 2
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
Case 2 n does not divide y (neither does it divide x) x^n + y^n = (e^n +x)^n also x^n + y^n = (f^n + y)^n Setting these equal and taking the nth root we get
- e^n + x = f^n + y
transposing we get
- e^n - f^n = y-x
which says the difference between y and x is a difference between nth powers.