Finding all possible roots of the equation

Question

Find all possible solutions to the equation $$(x^3-x)+(y^3-y)=z^3-z$$ where $(x,y,z)\gt1$ and $\in\mathbb{Z}$ and not all three of them are equal.

The original question didn't have the last condition of not all three of them are equal but I imposed this condition as, if all variables are equal then there exists infinite solutions.

My work:

We can easily see that permutations of $0,1$ are the trivial solutions of the equation. I don't know whether this equation has infinite solutions or not but I think so as, we can rewrite this equation as $$(x-1)(x)(x+1)+(y-1)(y)(y+1)=(z-1)(z)(z+1)$$ where each term is a product of three consecutive integers i.e. even if not all three of them are equal there may be infinite solutions.

Any help is greatly appreciated.

Answer 1

I was thinking like this (correct me if I am wrong):

map the function each of them:

$x^{3} - x → a^{2}$

$y^{3} - y → b^{2}$

$z^{3} - z → c^{2}$

By using Pythagorean' theorem:

$ a^{2} + b^{2} = c^{2}$

$ c = \sqrt{a^{2} + b^{2}}$

$ c = \sqrt{(x^{3} - x) + (y^{3} - y)}$

$ c = \sqrt{(x-1)(x)(x+1) - (y-1)(y)(y+1)}$

We can plot the graph of $c$ by using computer software like Julia or R or Python (with the boundary conditions such as $(x,y,z) > 1 ∧ x,y,z \in \mathbb{Z}$. When we know the value of $c$ we can easily plot all the possible values of $z$.

For the real analysis part, I am not an expert with this, but this is what I can think of: $(∀x, x \in X) ∧ (∀y, y \in Y) ↔ (x,y \in Z)$ with $x,y,z \in \mathbb{Z}$ and $(x,y,z) > 1$

Given the condition, we can safely assume that the solution set, $S$, is $(∀s, s \in S) ∧ (∀z, z \in Z): s \preceq z$

Answer 2

Using that: $$x^3+y^3=(x+y)(x^2-xy+y^2)$$ and similarly $$x^3-z^3=(x-z)(x^2+xz+z^2)$$ where $x$ and $y$ may be swapped, we arrive at three equations: $$x^3-x=(y-z)(y^2+yz+z^2-1)\tag 1$$ $$y^3-y=(x-z)(x^2+xz+z^2-1)\tag 2$$$$z^3-z=(x+y)(x^2-xy+y^2-1)\tag 3$$

Taking parity into account, we quickly see, since both $t^3-t$ and $t^3+u^3$ are always even, that $x,y,z$ must all be even.

I'm still working on the rest, but this is a good start

Answer 3

$$ 4^3 - 4 + 4^3 - 4 = 5^3 - 5 $$

$$ 9^3 - 9 + 15^3 - 15 = 16^3 - 16 $$

jagy@gost:~/Desktop/Cplusplus$ ./mse
    4    4    5   diff:  1
    9   15   16   diff:  1
   21   55   56   diff:  1
   31   56   59   diff:  3
   40   71   75   diff:  4
   62  103  110   diff:  7
   85   91  111   diff:  20
   35  119  120   diff:  1
   49  139  141   diff:  2
  120  155  176   diff:  21
  188  202  246   diff:  44
  101  291  295   diff:  4
  328  337  419   diff:  82
  150  430  436   diff:  6
  253  425  453   diff:  28
  249  451  475   diff:  24
  363  416  493   diff:  77
  220  516  529   diff:  13
  137  533  536   diff:  3
  425  449  551   diff:  102
  315  528  563   diff:  35
  435  496  589   diff:  93
  400  589  645   diff:  56
  325  664  689   diff:  25
  273  689  703   diff:  14
  305  696  715   diff:  19
  350  714  741   diff:  27
  533  644  748   diff:  104
  425  706  754   diff:  48
  379  791  819   diff:  28
  609  755  869   diff:  114
  231  904  909   diff:  5
  490  870  919   diff:  49
  776  951 1099   diff:  148
  704 1065 1159   diff:  94
  879 1045 1221   diff:  176
  969 1002 1242   diff:  240
  923 1287 1429   diff:  142
  291 1431 1435   diff:  4
  368 1437 1445   diff:  8
  856 1344 1451   diff:  107
  505 1630 1646   diff:  16
  898 1622 1709   diff:  87
  751 1691 1739   diff:  48
 1352 1479 1787   diff:  308