On x^3 - x + y^3 - y = z^3 - z
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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