Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.2 9/18/84; site dartvax.UUCP Path: utzoo!watmath!clyde!bonnie!akgua!whuxlm!harpo!decvax!dartvax!chuck From: chuck@dartvax.UUCP (Chuck Simmons) Newsgroups: net.math,net.puzzle Subject: Re: needed:algorithm Message-ID: <2980@dartvax.UUCP> Date: Sat, 27-Apr-85 16:13:03 EDT Article-I.D.: dartvax.2980 Posted: Sat Apr 27 16:13:03 1985 Date-Received: Mon, 29-Apr-85 00:14:23 EDT References: <1418@aecom.UUCP> <288@faron.UUCP> <7204@watdaisy.UUCP> <290@faron.UUCP> Distribution: net Organization: Dartmouth College, Hanover, NH Lines: 24 Xref: watmath net.math:1962 net.puzzle:797 > > >> Find the smallest integer which can be broken up into: > > >> a^4 + b^4 = k > > >> c^4 + d^4 = k > > > > > >I'll start by giving you a hint: since it's the sum of two fourth powers > > >and must be odd it is necessarily of the form 8n+1. > > > > The reasoning behind this conclusion escapes me. Why must it > > be odd? > > It's fairly simple: assume k is even. Then we have one of two cases either > a and b must be even or a and b must be odd. Either assumption leads to > a simple algebraic demonstration by infinite descent that there must be > a smaller solution. Call me dense, but this little "hint" doesn't do anything for me. I see 3 cases: clearly a = b mod 2 and c = d mod 2. So the 3 cases are a even, c even; a odd, c odd; a even, c odd. I can handle the first of these cases. But for the rest, I remain clueless. Could you provide a slightly more elaborate hint? -- Chuck Simmons