Do there exist 5 positive integers such that the product of any two is one less than a square? For a discussion of this and related questions, see If ab+1, ac+1, bc+1 are squares,... . As an aside, there's a connection between this problem and the sets of points called "clouds" by L. Comtet in his book "Advanced Combinatorics". If we have 3 (distinct) positive integers a,b,c (e.g., 1,3,8) such that each of the three pairwise products ab,ac,ad is a "shy square" (defined as one less than a square), then obviously the square (abc)^2 is expressible as a product of three shy squares, i.e., (abc)^2 = (A^2-1)(B^2-1)(C^2-1). However, if we have 4 distinct integers a,b,c,d (e.g., 1,3,8,120) such that each of the SIX pairwise products is a shy square, then the square (abcd)^2 must not only be expressible as a product of four shy squares, it must be so expressible in THREE distinct ways, corresponding to the three ways of choosing four out of the six pairwise products such that each of a,b,c,d appears exactly twice. For example, with a=1,b=3,c=8,d=120 we have (2880)^2 = (2^2 - 1)(3^2 - 1)(19^2 - 1)(31^2 - 1) = (2^2 - 1)(5^2 - 1)(11^2 - 1)(31^2 - 1) = (3^2 - 1)(5^2 - 1)(11^2 - 1)(19^2 - 1) Now, if we imagine 5 distinct integers a,b,c,d,e such that each of the TEN pairwise products is a shy square, then the square (abcde)^2 must be expressible as a product of 5 distinct shy squares in TWELVE distinct ways. It isn't trivial to find numbers that are expressible as products of shy squares in multiple ways. For example, no one has ever found ANY integer (let alone a square) that is expressible in the form (n^2 - 1)(m^2 - 1) in more than five distinct ways. (And only 6 five-way expressible numbers are known). Anyway, the connection to Comtet's "clouds" is that if you go on to consider N distinct integers such that each of the N(N-1)/2 pairwise products is a shy square, then the square of the product of those N integers must be expressible as a product of N shy squares in c(N) distinct ways, where the first few values of c(N) are N c(N) --- ----- 3 1 4 3 5 12 6 70 7 465 8 3507 This is the number of ways of choosing N out of N(N-1)/2 pairwise products of N numbers such that each of the individual numbers appears exactly twice. Checking the Encyclopedia of Integer Sequences (Sloane and Plouffe) we find that this is sequence M2937, which Comtet defined in terms of the intersection points of N lines in general position in the plane. Incidentally the question about 5 positive integers such that the product of any two is one less than a square goes back to Diophantus. It's true that Diophantus only required rational numbers, rather than restricting his domain to integers, but of course many of his results have relevance for integers too. In Book III of The Arithmetica he treated the problem of finding three numbers such that the product of any two of them increased by 1 is a square, and he gave the triple a=x, b=x+2, c=4x+4. Then in Problem 20, Book IV, he treated the problem of finding FOUR numbers such that all six pairwise products are 1 less than a square. He built on his triple solution a,b,c, but oddly enough he *didn't* use the case x=1 (as Fermat later did to find 1,3,8,120). Instead he decided to make the product of the first and fourth numbers equal to one less than (3x+1)^2. I suppose he was just following the pattern ab+1 = (1x+1)^2 ac+1 = (2x+1)^2 bc+1 = (2x+3)^2 ad+1 = (3x+1)^2 bd+1 = ? cd+1 = (6x+5)^2 Notice that setting ad+1 = (3x+1)^2 forces d to be 9x+6, which automatically gives cd+1 = (6x+5)^2. Diophantus didn't comment much on this, but it's clearly systematic. For example, if he went on to a 5th number e such that ae+1 = (4x+1)^2 it would force e to be 16x+8, which automatically gives de+1 = (12x+7)^2, and so on. Anyway, all he needs to do now is find a (rational) value of x such that bd+1 = 9x^2 + 24x + 13 is a (rational) square. He reasons that if we assume bd+1 = (3x+k)^2 for some integer k then we will have (24-6k)x + (13-k^2) = 0, and so x can be any number of the form x = (k^2 - 13)/(24 - 6k). He chooses k=-4, which gives x=1/16, so his four numbers are 1/16, 33/16, 68/16, and 105/16. (He might have chosen k=-11 to give x=6/5.) Euler gave a family of such rational quintuples, including these two examples 1/2 5/2 6 48 44880/128881 1 3 8 120 777480/8288641

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