Path: ...!news.misty.com!weretis.net!feeder9.news.weretis.net!panix!.POSTED.panix1.panix.com!not-for-mail From: "Keith F. Lynch" Newsgroups: rec.puzzles Subject: Re: Three rational triples Date: Thu, 19 Sep 2024 12:16:57 -0000 (UTC) Organization: United Individualist Message-ID: References: <60e28add10c8ff75a90697397ee0ba6f@www.novabbs.com> Injection-Date: Thu, 19 Sep 2024 12:16:57 -0000 (UTC) Injection-Info: reader1.panix.com; posting-host="panix1.panix.com:166.84.1.1"; logging-data="703"; mail-complaints-to="abuse@panix.com" X-Newsreader: trn 4.0-test77 (Sep 1, 2010) Bytes: 1898 Lines: 32 HenHanna wrote: > Keith F. Lynch wrote: >> Since it's been more than a week, and nobody has figured it out: >> Each of them has a sum that's equal to its product and is an integer. > i think one person said exactly that. Who and when? I didn't see any such post. > Is it easy to find them? No, even though there are infinitely many. Try and find one I didn't list. Constraints: All three numbers must be positive, real, and rational, but not integers. And of course have to be in simplest form, i.e. 1/2, not 2/4. One person posted "52/39, 7/6, 9/2" which is of course the same three numbers as my "9/2, 4/3, 7/6". And he didn't say what property they had, anyway. Without any constraints, x,i,-i is always a solution for any x, if i is the square root of minus one. > How about 2 numbers Too simple to be interesting. If you want two numbers to have a sum of S and a product of P, whether or not S=P, the two numbers will be S + sqrt(S^2 - 4P) and S - sqrt(S^2 - 4P). -- Keith F. Lynch - http://keithlynch.net/ Please see http://keithlynch.net/email.html before emailing me.