Suppose we wish to find an infinite set of solutions of the equation x^3 - x + y^3 - y = z^3 - z (1) where x, y, z are integers greater than 1. If z and x are both odd or both even, we can define integers u and v such that z=u+v and x=u-v. Substituting into equation (1) gives y^3 - y = 2v(3u^2 + v^2 - 1) Since v divides the right-hand side, it would be nice if it also divides the left-hand side, which can be written as (y)(y^2 - 1). By setting y = kv for some integer k we can divide through the entire equation by v, leaving just an equation of degree 2 in the variables u and v. Making this substitution and dividing through by v, we have k(k^2 v^2 - 1) = 6u^2 + 2v^2 - 2 Re-arranging terms, this can be written in the form of a Pell equation in the variables u and v for any given k (k^3 - 2)v^2 - 6u^2 = (k - 2) With k equal to 1 or 2, this equation has either no solutions or else trivial solutions. However, with k = 3 we have (5v)^2 - 6u^2 = 1 This Pell equation has the infinite family of solutions u 5v ------- ------- 2 5 20 49 198 485 1960 4801 19402 47525 192060 470449 1901198 4656965 etc. Each of these sequences satisfies the recurrence s[n] = 10s[n-1] - s[n-2] The sequence for 5v has the initial values s[0]=5 and s[1]=49, and we can combine successive recurrences to give a single recurrence for just the even-indexed terms s[2n] = 98s[2(n-1)] - s[2(n-2)] with the initial values 5 and 485. Hence all subsequent even- indexed terms of the 5v sequence are divisible by 5. Thus we have the infinite sequence of solutions u v x=u-v y=3v z=u+v ------- ------- ------- ------- ------- 2 1 1 3 3 198 97 101 291 295 19402 9505 9897 28515 28907 1901198 931393 969805 2794179 2832591 etc. Other infinite sequences of solutions can be constructed based on other values of k. However, as is characteristic of the solutions of Pell equations, these are exponentially increasing sequences, so they do not account for a very large fraction of all the solutions of equation (1). To find an efficient way of searching for more solutions, we can change the sign of z (without loss of generality) and write equation (1) in the equivalent form x^3 + y^3 + z^3 = x + y + z Making use of the algebraic identity x^3 + y^3 + z^3 = (x+y+z)^3 - 3(x+y)(x+z)(y+z) we have (x+y+z-1)(x+y+z)(x+y+z+1) = 3(x+y)(x+z)(y+z) (2) Letting S denote the sum x+y+z, the left side is S^3 - S, which is automatically divisible by 3 (due to Fermat's Little Theorem). So, we can search form solutions by considering each possible value of S in turn, and all the factorizations of (S^3 - S)/3 into three factors, which we can call a=x+y, b=x+z, and c=y+z. Then, for any such factorization, we can check to see if a+b+c = 2S. If it does, then we have the solution a+b-c a-b+c -a+b+c x = ----- y = ----- z = ------ 2 2 2 Of course, the signs of a,b,c ambiguous, because we can negate any two of them and leave the product abc unchanged. However, it's clear that (for all sufficiently large solutions) the two smaller factors (in magnitude) must have the same sign, and the larger factor must have the opposite sign, so we can simply take the smaller factors b,c as negative and a as positive. Aside from the degenerate "S=1" solutions of the form |x|=1,y=-z, the smallest solutions are shown in the table below. S c b a x y z ----- ----- ----- ----- ----- ----- ----- 1 1 1 0 0 0 1 2 1 1 2 1 1 0 2 1 2 1 1 0 1 3 -1 -1 8 4 4 -5 3 2 2 2 1 1 1 8 -1 -7 24 9 15 -16 20 -1 -35 76 21 55 -56 28 -3 -28 87 31 56 -59 34 -1 -85 154 35 119 -120 36 -4 -35 111 40 71 -75 47 -2 -92 188 49 139 -141 55 -7 -48 165 62 103 -110 65 -20 -26 176 85 91 -111 97 -4 -194 392 101 291 -295 99 -21 -56 275 120 155 -176 134 -3 -399 670 137 533 -536 144 -6 -286 580 150 430 -436 144 -44 -58 390 188 202 -246 207 -13 -309 736 220 516 -529 225 -24 -226 700 249 451 -475 225 -28 -200 678 253 425 -453 226 -5 -678 1135 231 904 -909 246 -82 -91 665 328 337 -419 259 -14 -430 962 273 689 -703 280 -35 -248 843 315 528 -563 286 -19 -410 1001 305 696 -715 286 -77 -130 779 363 416 -493 287 -4 -1144 1722 291 1431 -1435 300 -25 -364 989 325 664 -689 323 -27 -391 1064 350 714 -741 323 -102 -126 874 425 449 -551 342 -93 -154 931 435 496 -589 344 -56 -245 989 400 589 -645 351 -28 -440 1170 379 791 -819 360 -8 -1077 1805 368 1437 -1445 377 -48 -329 1131 425 706 -754 429 -104 -215 1177 533 644 -748 433 -6 -1732 2604 439 2165 -2171 441 -49 -429 1360 490 870 -919 489 -16 -1141 2135 505 1630 -1646 495 -114 -260 1364 609 755 -869 524 -5 -2615 3668 529 3139 -3144 610 -94 -455 1769 704 1065 -1159 628 -148 -323 1727 776 951 -1099 703 -48 -988 2442 751 1691 -1739 703 -176 -342 1924 879 1045 -1221 720 -10 -2876 4326 730 3596 -3606 729 -240 -273 1971 969 1002 -1242 736 -7 -3680 5159 743 4416 -4423 749 -50 -1070 2618 799 1819 -1869 749 -107 -595 2200 856 1344 -1451 781 -142 -506 2210 923 1287 -1429 811 -87 -811 2520 898 1622 -1709 813 -22 -2146 3794 835 2959 -2981 819 -35 -1599 3272 854 2418 -2453 863 -6 -5172 6904 869 6035 -6041 1024 -82 -1280 3410 1106 2304 -2386 1035 -55 -1739 3864 1090 2774 -2829 1035 -69 -1480 3619 1104 2515 -2584 1035 -105 -1081 3256 1140 2116 -2221 1036 -111 -1037 3220 1147 2073 -2184 1044 -308 -435 2831 1352 1479 -1787 1153 -8 -6918 9232 1161 8071 -8079 There are two distinct solutions for each of the S-values 3 = (3) 144 = (2^4)(3^2) 225 = (3^2)(5^2) 286 = (2)(11)(13) 323 = (17)(19) 703 = (19)(37) 749 = (7)(107) and there are three distinct solutions for the S-value 1035 = (3^2)(5)(23) It's also interesting to note the odd fact that, beginning with the 7th entry on this list, every 7th entry has a single-digit value of c, with the exception of the 21st entry. However, even in this case it is a near miss, because the 22nd entry has a single-digit c. Another item of interest is that in three of the cases where there are multiple solutions for a single S, there is also a solution for S+1. This is not too suprising, since the numbers to be factored consist of three consecutive integers S-1, S, and S+1, which implies that the candidates for consecutive S values share two factors. Noting that exactly one of the factors on the left side of (2) is divisible by 3, we can consider the most general factorization of the two sides. There are three cases to consider, depending on whether 3 divides S-1, S, or S+1. If 3 divides S, then we have integers A,B,..,I such that / x+y+z \ (x+y+z-1)( ----- )(x+y+z+1) = (x+y)(x+z)(y+z) \ 3 / (ABC) (DEF) (GHI) = (ADG)(BEH)(CFI) Therefore, we have ABC = x+y+z-1 3DEF = x+y+z GHI = x+y+z+1 ADG = x+y BEH = x+z CFI = y+z This immediately gives many relations between the factors A,B,I, such as 6DEF = ADG + BEH + CFI 3DEF = ABC + 1 = GHI - 1 These are necessary and sufficient conditions for a solution. Thus for any S divisible by 3 we need only check the 3-part factorizations S-1 = ABC S/3 = DEF S+1 = GHI to see if the integers A,B,...,I satisfy 6DEF=ADG+BEH+CFI. Of course, the orderings of the factors can be changed, so we can hold one of the orderings (say, ABC) fixed and apply all the permutations of the other two orderings. This gives 36 possible arrangements. Also, two of the numbers D,E,F could be negative, so if we are dealing with only positive numbers we must allow for the three possibilities 6DEF = ADG - BEH - CFI 6DEF = -ADG + BEH - CFI 6DEF = -ADG - BEH + CFI There is a solution for S if and only if one of the 36 permutations of the 3-part factorizations of S-1, S/3, S+1 satisfies one of these three relations. For example, with S = 99 we have 98 = (1)(14)(7) 99/3 = (11)(-1)(-3) 100 = (25)(4)(1) corresponding to the assignments A = 1 B = 14 C = 7 D = 11 E = -1 F = -3 G = 25 H = 4 I = 1 Computing the quantity ADG+BEH+CFI gives (1)(11)(25) + (14)(-1)(4) + (7)(-3)(1) = 198 which equals 6DEF. Hence we have the solution x = ( ADG + GEH - CFI)/2 = 120 y = ( ADG - GEH + CFI)/2 = 155 z = (-ADG + GEH + CFI)/2 = -176

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