Wed, 29 Oct 2003 17:28:31 -0500 (EST) Dan Hoey <Hoey@aic.nrl.navy.mil> Hmm, still pretty dumb. Suppose we also require that a,b,c,d are four different positive integers? I've gotten b and d different, but I haven't managed all four. I think I did, when I sent mail to J. Perry this morning (appended below). The unique integer question was one possible interpretation, but it, too, didn't seem very challenging. In case that's still dumb, can all of a,b,c,d be made relatively prime in pairs? Here's my note from this morning, re: unique a,b,c,d,x,y. To: "Jon Perry" <perry@globalnet.co.uk> Cc: greenwald@central.cis.upenn.edu, "Christian G. Bower" <bowerc@usa.net> Subject: Re: [math-fun] Simple EF In-reply-to: Your message of "Wed, 29 Oct 2003 16:43:40 GMT." <MBBBLALCNBNKDIOOGHIOIEPKFDAA.perry@globalnet.co.uk> Date: Wed, 29 Oct 2003 13:51:07 EST From: Michael B Greenwald <mbgreen@central.cis.upenn.edu> Wed, 29 Oct 2003 16:43:40 -0000 "Jon Perry" <perry@globalnet.co.uk> This may be completely dumb, but is it always possible to solve x/y = a/b + c/d, where all algebraic quantities are positive integers, for all x and y? [I wasn't sure exactly what you are asking, so I removed math-fun to avoid filling up everyone's mailbox with useless verbiage.] I assume you are not asking for the trivial answer. That is, c = (xb - ay), and d = by; ... so if you choose q = a/b with q < x/y, then c/d is positive. ... So, either you are asking for x/y = a/b + c/d where a,b,c,d,x,y are all unique, or else you are asking when x/y aren't integers and x/y is irrational, or else ?? In the first case, if you are looking for unique values for x,y,a,b,c,d then it is always possible in many ways. For one example (maybe not the simplest), choose a to be the smallest prime relatively prime to both x and y, and choose b to be any prime greater than axy. Then d = by, which is clearly unequal to x,y,a, or b. c = (xb - ay) can't be equal to any of a,b,x or y, because x and b are each relatively prime to both a and y, so their difference can't share any prime factors with them (let alone be equal to them). In the second case, if x and y are irrational then it is clearly not always possible to solve your equation above, because (ad + bc)/bd is rational, and therefore can never = x/y if x/y is irrational. You may be asking for something else (the "??" case), but I can't figure out what it is, though.