tag:blogger.com,1999:blog-5871184605628305545.post8801527431129486523..comments2013-05-25T03:36:33.414-07:00Comments on Puzzled: Total SolutionsAnonymoushttp://www.blogger.com/profile/02364548875097103884noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-5871184605628305545.post-22793651307465171812012-08-10T11:10:58.511-07:002012-08-10T11:10:58.511-07:00@rik - The question is too find the total number o...@rik - The question is too find the total number of solutions rather than a particular solution<br /><br />Soln----<br /><br />1/x + 1/y = 1/k<br />On simplifying we get<br />(x-k)(y-k) = k^2<br />Now, the total solutions would be nothing but the total factors of k^2 which can be easily find out.Anonymoushttps://www.blogger.com/profile/02364548875097103884noreply@blogger.comtag:blogger.com,1999:blog-5871184605628305545.post-6128863102222516822012-07-15T00:46:26.091-07:002012-07-15T00:46:26.091-07:00^in the above argument, p can be any divisor of k,...^in the above argument, p can be any divisor of k, it need not be necessary that gcd(p,q)=1 and the solution remains the same....Rikhttps://www.blogger.com/profile/16765576918216443372noreply@blogger.comtag:blogger.com,1999:blog-5871184605628305545.post-73334546527908939332012-07-14T13:35:27.307-07:002012-07-14T13:35:27.307-07:00find any p,q s.t. k=p*q and gcd(p,q)=1... this can...find any p,q s.t. k=p*q and gcd(p,q)=1... this can be done using fundamental theorem of arithmetic... k is uniquely representable by its prime factorization and p&q are partition of prime factorization of k i.e. p,q have different primes as their divisor and hence no common divisor.... now the solution is symmetric about x,y and the solution is x =(k+p), y = k*(k/p + 1) .... (note k/p = q, an integer)...Rikhttps://www.blogger.com/profile/16765576918216443372noreply@blogger.com