Problem: To find all real solutions of the system: 3 a = ( b + c + d ) 3 3 b = ( c + d + e ) 3 3 c = ( d + e + a ) 3 3 d = ( e + a + b ) 3 3 e = ( a + b + c ) 3

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Problem: To find all real solutions of the system:  3  a  =  (  b  +  c  +  d      )    3    3  b  =  (  c  +  d  +  e      )    3    3  c  =  (  d  +  e  +  a      )    3    3  d  =  (  e  +  a  +  b      )    3    3  e  =  (  a  +  b  +  c      )    3

hoffwnbu · Accepted Answer

A real solution must satisfy   a  =  b  =  c  =  d  =  e. To see this, let an arbitrary solution   (  a  ,  b  ,  c  ,  d  ,  e  ) be given, and replacing it with a cyclic shift, we can assume without loss of generality that   (  a  ,  b  ,  c  ,  d  ,  e  ). Then   a  ≥  e, so that             3      a        3    ≥            3      e        3  . Using the equations   3  a  =  (  b  +  c  +  d      )    3   and   3  e  =  (  a  +  b  +  c      )    3  , this implies that   b  +  c  +  d  ≥  a  +  b  +  c, i.e.,   d  ≥  a. But since   a  =  max  {  a  ,  b  ,  c  ,  d  ,  e  }  ≥  d, this implies   d  =  a. Iterating this same chain of reasoning with   d in place of   a, we obtain, in turn,   b  =  d,   e  =  b, and   c  =  e, so that indeed   a  =  b  =  c  =  d  =  e.It follows that the only real solutions satisfy   3  a  =  (  b  +  c  +  d      )    3    =  (  3  a      )    3  , so that   9      a    3    =  a, which has only solutions   a  ∈  {  0  ,  ±      1    3    }. Therefore, the real solutions are   (  0  ,  0  ,  0  ,  0  ,  0  ),   (      1    3    ,      1    3    ,      1    3    ,      1    3    ,      1    3    ), and   (  −      1    3    ,  −      1    3    ,  −      1    3    ,  −      1    3    ,  −      1    3    ).

Problem: To find all real solutions of the system: 3 a = ( b + c + d

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