A nonlinear system of equations { y 3 − 9 x 2 + 27 x − 27 = 0 z 3 − 9 y 2 + 27 y − 27 = 0 x 3 − 9 z 2 + 27 z − 27 = 0

Question

A nonlinear system of equations      {                                        y            3                    −          9                      x            2                    +          27          x          −          27                          =                          0                                                  z            3                    −          9                      y            2                    +          27          y          −          27                          =                          0                                                  x            3                    −          9                      z            2                    +          27          z          −          27                          =                          0

livin4him777lf · Accepted Answer

For integer solutions:You can rewrite the equations using      y    3    −  9      x    2    +  27  x  −  27  =      y    3    −      x    3    +  (  x  −  3      )    3  and similarly for the two other equations. Adding the three equations then yields  (  x  −  3      )    3    +  (  y  −  3      )    3    +  (  z  −  3      )    3    =  0.It is long known that in all integer solutions of the equation       a    3    +      b    3    =      c    3  , one of the integers must be   0. Our equation is (modulo rearranging) exactly that, so we must have   x  =  3 or   y  =  3 or   z  =  3.Suppose   x  =  3. Then the first equation reduces to       y    3    −      x    3    =  0, hence       y    3    =  27, hence   y  =  3. In the same way, it follows that   z  =  3. Starting from   y  =  3 or   z  =  3 yields the same. So   (  3  ,  3  ,  3  ) is the only integer solution to the system.Now, for any real solution, we still have by adding the equations  (  x  −  3      )    3    +  (  y  −  3      )    3    +  (  z  −  3      )    3    =  0  ,so if one of   x  ,  y  ,  z were different from   3, one of the three at least would be larger than   3. Without loss of generality, let   x  =  3  +  δ  &amp;gt;  3. Then, from the first equation, we obtain                              y          3                                    =                  x          3                −        (        x        −        3                  )          3                                                  =        (        3        +        δ                  )          3                −                  δ          3                                                  =                  3          3                +        27        δ        +        9                  δ          2                                                  &amp;gt;                  3          3                ,            so also   y  &amp;gt;  3. The same reasoning for the second equation then yields   z  &amp;gt;  3, and so  (  x  −  3      )    3    +  (  y  −  3      )    3    +  (  z  −  3      )    3    &amp;gt;  0contradicting the assumption that   (  x  ,  y  ,  z  ) is a solution of the system.Hence   (  3  ,  3  ,  3  ) is the only real solution of the system.

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