Newsgroups: sci.math From: tycchow@math.mit.edu (Timothy Y. Chow) Subject: Re: Help!! - Problem to solve Date: Tue, 28 Mar 95 19:04:23 GMT In article <1995Mar27.214349.88740@kuhub.cc.ukans.edu>, Daniel M. O'Brien wrote: > Anyway - I noticed that the product of the seeds on either side > of the bracket is the sum of the seeds on the other side. My question > is - are there any four numbers besides 1,2,3 and 5 for which this > condition is true?? Nice question! The problem is to find all integer solutions to the simultaneous equations ab = c + d cd = a + b. Let us first consider the case where a and b are both (strictly) negative. Then c + d = ab > 0 so either c or d is positive, and cd = a + b < 0 so exactly one of c and d is positive and the other is negative---say c > 0 and d < 0. Let w = -a, x = -b, y = c, and z = -d, so that w, x, y and z are all positive. Then wx = y - z yz = w + x, so wx < y <= yz = w+x, and since w and x are positive integers, wx < w+x implies that either w or x equals 1. Without loss of generality we may assume x = 1. Then yz = w + 1 so y divides w + 1, but w = y - z so y > w. Hence y = w + 1 and z = 1. Thus (a,b,c,d) = (-n,-1,n+1,-1) is a solution for any positive integer n, and these are the only solutions for which a and b are both negative (up to interchange of a and b, or of c and d). So we may now assume without loss of generality that a is nonnegative. From our original two equations it follows easily that (a-1)(b-1) + (c-1)(d-1) = 2, so either (a-1)(b-1) or (c-1)(d-1) is positive. Again without loss of generality we assume that (a-1)(b-1) > 0. If b is negative, this forces a = 0 (since a is nonnegative) and hence c = -d and b = -d^2. Hence we have the family of solutions (a,b,c,d) = (0,-n^2,-n,n) for any integer n. Otherwise, both a and b are nonnegative. Then c + d = ab >= 0 so either c or d is nonnegative. Now cd = a + b >= 0, so if one of c or d (say c) is negative then this forces d = 0 and hence a = b = 0, but a = b = d = 0 forces c = 0, a contradiction. Hence both c and d are nonnegative. Now if either c or d is zero, then a + b = cd = 0 so a = b = 0 and hence d = 0, giving us the solution (a,b,c,d) = (0,0,0,0). So let us assume now that both c and d are strictly positive. Then (c-1)(d-1) is a nonnegative integer and (from an earlier assumption) (a-1)(b-1) is a positive integer, so from (a-1)(b-1) + (c-1)(d-1) = 2, we see that we just need to check the cases (a-1)(b-1) = (c-1)(d-1) = 1 and (a-1)(b-1) = 2, (c-1)(d-1) = 0. A quick exhaustive search shows that the only new solutions obtained are (2,2,2,2) and (2,3,1,5). Summary: up to the interchanges a <-> b, c <-> d, and (a,b) <-> (c,d), the set of solutions is {(2,2,2,2), (2,3,1,5), (-n,-1,n+1,-1), (0,-n^2,-n,n)}, where n is any integer. -- Tim Chow tycchow@math.mit.edu Where a calculator on the ENIAC is equipped with 18,000 vacuum tubes and weighs 30 tons, computers in the future may have only 1,000 vacuum tubes and weigh only 1 1/2 tons. ---Popular Mechanics, March 1949