Playing with Fermat's Last Theorem
From RoyalWeb
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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, | + | by Will Johnson, [mailto:wjhonson@aol.com wjhonson@aol.com] copyright 2001-4, all rights reserved. |
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(Condition 1) x^n + y^n = z^n | (Condition 1) x^n + y^n = z^n | ||
| − | (Condition 2) n prime and not 2 | + | |
| − | (Condition 3) x,y,z natural numbers | + | (Condition 2) n is prime and is not 2 |
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| + | (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 | + | 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. |
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| − | From Condition 3 and Condition 2 | + | 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) |
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| − | Since d divides every term on the right, it must also evenly | + | Since d divides every term on the right, it must also evenly 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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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 | |
| − | (Result 5) each p(j) = y(o) for some o and | + | (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 | + | |
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) | ||
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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 | + | 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. |
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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 | + | 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. |
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| − | divisible by n^n. Trying each successive n^q one by one we | + | (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 | + | p(j) = n AND l(j) = t(o)n - 1 |
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OR l(j) = t(o)n | OR l(j) = t(o)n | ||
| − | Now this is true for every element of the prime factorization | + | 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 | |
| − | (Result 9) if n divides d the other elements of its prime factorization must be nth | + | |
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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 | |
| − | if n does not divide d all elements of d's prime factorization must be nth powers so | + | 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 |
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| − | Now we arbitrarily chose y to act upon, so therefore choosing | + | Now we arbitrarily chose y to act upon, so therefore choosing x has the same result. |
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| − | Reducing equation 5, mod n, we get y^n congruent to d^n mod n | + | 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 | + | The case n divides y is analogous to the case n divides x 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 | ||
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| − | + | 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 | (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 | + | 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 | + | 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 |
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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 | ||
