tag:blogger.com,1999:blog-3722233.post7539655553907516201..comments2024-10-10T06:29:39.038-05:00Comments on Computational Complexity: Is there an easier proof? A less messy proof? Lance Fortnowhttp://www.blogger.com/profile/06752030912874378610noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-3722233.post-10828554298944965042015-07-17T18:49:43.224-05:002015-07-17T18:49:43.224-05:00I tried a little bit and came up with this.
(1)
x...I tried a little bit and came up with this.<br /><br />(1)<br />x³+y³+z³ = (x²+y²+z²) - (xy+xz+yz)*(x+y+z) + 3xyz<br />c = b*a - (xy+xz+yz)*a + 3xyz<br />substituting: (xy+xz+yz) = (a² - b) / 2<br /><br />we get a formula xyz = ... only a,b,c<br /><br />(2)<br />(y+z)² = (a-x)²<br />y² + 2yz + z² = (a-x)²<br />b - x² + 2yz = (a-x)²<br /><br />now substitute yz with (1) to get a term of the form 1/x. This will lead to a polynomial with x of third degree in terms of a,b,c.<br /><br />(4)<br />--> we have at least one real solution and at most 3<br />--> use the discriminant to show that 3 solution is not possible. (I did not do that)<br /><br />BenjaminAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-3722233.post-74611232964419688952015-07-14T18:54:26.153-05:002015-07-14T18:54:26.153-05:00If there ARE two solutions, say (x,y,z) and (u,v,w...If there ARE two solutions, say (x,y,z) and (u,v,w), then (x-u,y-v,z-w) is a singular vector for the matrix with columns ( 1 // x + u // x² + xu + u²) etc.; which means that matrix must be singular; the permutations evidently provide solutions (e.g, the tanspositions induce repeated columns), ... and... I don't know what to suggest next.<br /><br />On the Other Side of Things, the mod k classes of the e_j for j < k, with the kth power sum do not determine the mod k class of e_k; so any argument that works for the finite rings will have to mention large primes.Belfry Bathttps://www.blogger.com/profile/00514867101036143597noreply@blogger.com