Solving a^n + b^n = c^nDate: 03/10/2009 at 22:00:13 From: Tim Subject: Fermat's last theorem What if n < 2 ? If n = -2, we have c^2 = (a^2*b^2)/(a^2 + b^2) and there is a solution. What is the general approach? Date: 03/13/2009 at 14:26:59 From: Doctor Vogler Subject: Re: Fermat's last theorem Hi Tim, Thanks for writing to Dr. Math. When n is negative, say n = -m, then c^(-m) = a^(-m) + b^(-m) is equivalent (when abc is nonzero) to c^(-m) = (a^m + b^m)/(ab)^m c^m = (ab)^m/(a^m + b^m) c^m(a^m + b^m) = (ab)^m (ac)^m + (bc)^m = (ab)^m and so by Fermat's Last Theorem, there are no integer solutions when m is bigger than 2. So the only cases left to check are n = -2, -1, 0, 1, and 2. Of course, for n=2, we have the well-known Pythagorean triples: Dr. Math FAQ: Pythagorean Triples http://mathforum.org/dr.math/faq/faq.pythag.triples.html When n=1, the solution is easy, since c = a + b implies that all solutions have the form (a, b, a+b). When n=0, there is no solution, because the equation becomes 1 = 2. When n=-1, the solution is a little trickier, but not too bad. You end up with ac + bc = ab, which you can also factor as (a-c)(b-c) = c^2. So we could let r = a-c and let s = b-c, and then rs = c^2. If we let g = gcd(r, s), then (r/g)(s/g) = (c/g)^2, and r/g and s/g are relatively prime integers whose product is a square. So that means that both r/g and s/g are squares of integers, which we will call r/g = u^2 and s/g = v^2. And then c/g = uv. Therefore, all solutions come from a relatively prime pair (u, v) and an integer g, giving the solution (gu(u+v), gv(u+v), guv). Finally, we have the case that you asked about, where n = -2 and c^2 = (ab)^2/(a^2 + b^2) or (ac)^2 + (bc)^2 = (ab)^2. You can expect this is a little bit harder than the case of n=2. If we use the method of Rational Solutions to Two Variable Quadratic Equation http://mathforum.org/library/drmath/view/65319.html then it is pretty easy to determine that all rational solutions have the form a = 2t(t^2 + 1)k b = (t^4 - 1)k c = 2t(t^2 - 1)k for some rational numbers t and k. So all integer solutions have the form a = 2uv(u^2 + v^2)k b = (u^4 - v^4)k c = 2uv(u^2 - v^2)k where u and v are relatively prime integers and k is a rational number. But you can show that gcd(2uv(u^2 + v^2), u^4 - v^4, 2uv(u^2 - v^2)) is either 4 (if u and v are both odd) or 1. Therefore, all integer solutions have the form a = 2uv(u^2 + v^2)k/4 b = (u^4 - v^4)k/4 c = 2uv(u^2 - v^2)k/4 for some pair of relatively prime *odd* integers (u, v), and some integer k, or a = 2uv(u^2 + v^2)k b = (u^4 - v^4)k c = 2uv(u^2 - v^2)k for some pair of relatively prime integers (u, v), not both odd, and some integer k. You could probably also do a trick like for Pythagorean triples where you eliminate the other case by switching a and b. That is, if we have a = 2uv(u^2 + v^2)k/4 b = (u^4 - v^4)k/4 c = 2uv(u^2 - v^2)k/4 with u and v both odd, then we can simply set U = (u+v)/2 and V = (u-v)/2 so that b = 2UV(U^2 + V^2)k a = (U^4 - V^4)k c = 2UV(U^2 - V^2)k We could probably get at the same result by factoring the equation as a^2*c^2 = b^2(a - c)(a + c), assuming some coprimality condition, and building up the solution like in the link above on Pythagorean triples. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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