Diophantus posed the problem of finding three rational numbers a,b,c such that ab+1, ac+1 and bc+1 are each squares, and this problem has been extended and generalized (and specialized) in several different ways by later mathematicians. For example, Euler found sets of four, and one set of FIVE, rationals numbers such that each of the pairwise products is one less than a square. A discussion of this general problem can be found in If ab+1, ac+1, bc+1 are squares.... There's also an article in the Feb 98 Mathematics Magazine on this subject. Anyway, John Gowland recently proposed a slightly different extension of this problem, namely, to find three integers such that all FOUR of the quantities ab+1, ac+1, bc+1, abc+1 are distinct squares, i.e., we require integer solutions of the simultaneous equations ab+1 = C^2 ac+1 = B^2 bc+1 = A^2 abc+1 = D^2 Obviously any such triple is a solution to the original problem of Diophantus, but the 4th condition involving the product of all three numbers significantly restricts the set of solutions. Notice that we stipulated "distinct" squares, because some of the infinitely many solutions of Diophantus's original problem have a=1, and since bc+1 is a square they automatically satisfy abc+1=square as well. To avoid these trivial solutions we require all four of the squares to be distinct. (Equivalently, each of the numbers a,b,c must be greater than 1.) To characterize the solutions of this extended problem, it might be useful to recall Saunderson's parameterization giving a large class of solutions a,b,c of the original 3-square problem: a = n b = q(qn+2) c = (q+1)[(q+1)n+2] where n is an integer and q is any RATIONAL number such that b and c are integers. For example, with n=45 and q=2/5 we have a=45 b=8 c=91 This example wan't chosen entirely at random, because it happens to also be a solution of the extended problem, as shown by the fact that (45)(8)(91) + 1 = 181^2 However, not surprisingly, only a small fraction of the solutions given by Saunderson's parameterization are also solutions of the extended problem. Notice that if we substitute Saunderson's expressions for a,b,c into the equation abc+1 = D^2 we have a quartic in q which happens to be fairly easy to solve explicitly, giving the result __________________________________ -(n+2) +- / n^2 + 4 +- 4 sqrt(n(n + D^2 - 1)) q = --------------------------------------------- 2n Now, we need q to be rational, so the quantity under the radical must be a square. Thus we have an integer w such that w^2 = n^2 + 4 +- sqrt(n(n + D^2 - 1)) and this requires that the quantity in the square root must be a square, so we have an integer m such that m^2 = n^2 + n(D^2 - 1) Solving this for n leads us to conclude that (D^2-1)^2 plus 4m^2 must be a square, so we have a Pythagorean triple (D^2 - 1)^2 + (2m)^2 = R^2 for some integer R. Obviously this implies the existence of an integer k and mutually coprime integers x,y (one odd, one even) such that D^2 - 1 = 2kxy 2m = k(x^2 - y^2) R = k(x^2 + y^2) or possibly swapping the first two expressions depending on whether 2m/k is odd or even. Here's a little table summarizing the smallest solutions of the full 4-square problem a b c D k x y --- --- --- ----- ----- ----- ----- 5 7 24 29 10 7 6 45 8 91 181 90 14 13 84 20 186 559 168 31 30 102 44 280 1121 204 56 55 105 8 171 379 210 19 18 119 40 297 1189 238 55 54 133 3 176 265 266 12 11 105 11 184 461 210 23 22 301 24 495 1891 602 55 54 Several regularities are apparent in this table, e.g., the value of k is always equal to 2a, and we can verify that m always equals a(x+y), which is obvious because x-y = 1 in all cases. If we take advantage of these regularities, the problem reduces to finding integers n and x such that D^2 = 4nx(x-1) + 1 (1) w^2 = n^2 + 4 +- 4n(2x-1) Given such integers, the value of Saunderson's q is then n + 2 +- w q = - ---------- 2n from which we can compute the values of the triple a,b,c. Here is a table of all the solutions given by this form for n and x less than 10000: n x a b c A B C D ---- ---- ----- ----- ----- ----- ----- ----- ------- 5 7 5 7 24 13 11 6 29 45 14 8 45 91 64 27 19 181 3 77 3 133 176 153 23 20 265 105 19 8 105 171 134 37 29 379 11 70 11 105 184 139 45 34 461 20 63 20 84 186 125 61 41 559 44 85 44 102 280 169 111 67 1121 119 55 40 119 297 188 109 69 1189 301 55 24 301 495 386 109 85 1891 477 66 24 477 715 584 131 107 2861 85 456 85 672 1235 911 324 239 8399 114 561 114 816 1540 1121 419 305 11969 165 498 165 664 1491 995 496 331 12781 132 575 132 820 1610 1149 461 329 13201 280 469 280 546 1608 937 671 391 15679 2387 175 40 2387 3045 2696 349 309 17051 74 2065 74 3612 4720 4129 591 517 35519 85 6432 85 11859 13952 12863 1089 1004 118591 120 9880 120 18278 21360 19759 1601 1481 216449 3193 4182 3193 4551 15368 8363 7005 3812 472565 5705 7258 5705 7831 26904 14515 12389 6684 1096339 An exhaustive search of all integer triples a,b,c satisfying all four conditions doesn't seem to turn up any additional solutions (in this range) beyond those listed here, so it's possible that (1) represents the necessary and sufficient conditions. Of course, there are many symmetries here, beyond what I've explicitly used, and I've just listed the first (n,x) yielding each solution triple, but there are always at least three such pairs that give the same triple (corresponding to the fact that we can identify the parameter n with a, b, or c). One striking feature of the table is how, when arranged by increasing D, the values of n occur in "sawtooth" cycles, with the following values 5 45 3 105 11 20 44 119 301 477 85 114 165 132 280 2387 74 85 120 3193 5705 ... I suppose this might just be an artifact of that way I've selected representative solutions and arranged them, but it seems to suggest some underlying pattern that hasn't been brought out explicitly. By the way, this problem also relates to the subject of representing numbers as products of "shy squares" in multiple ways. (A "shy square is one less than a square). Solutions to the general problem above obviously represent integers that can be factored in two distinct ways into shy squares, namely (A^2 - 1)(B^2 - 1)(C^2 - 1) = (D^2 - 1)^2 In other words, this is a product of three distinct shy squares that is also the square of a shy square. There are several open questions in this area. For example, it's possible for a number to be the product of two shy squares in 5 distinct ways, but it's not known if there exists a six-way expressible number.

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