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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