Find all triplets ( x , y , z ) that satisfy: x 2012 + y 2012 + z 2012 = 3 x 2013 + y 2013 + z 2013 = 3 x 2014 + y 2014 + z 2014 = 3Trivial solution: ( 1 , 1 , 1 )

Question

Find all triplets   (  x  ,  y  ,  z  ) that satisfy:      x          2012        +      y          2012        +      z          2012        =  3        x          2013        +      y          2013        +      z          2013        =  3        x          2014        +      y          2014        +      z          2014        =  3Trivial solution:   (  1  ,  1  ,  1  )

Amiya Watkins · Accepted Answer

Let       e    1  ,       e    2   and       e    3   be LHS of the the first, second and third equations respectivelly, then:      e    3    −  2      e    2    +      e    1    =        x          2012        (      x    2    −  2  x  +  1  )  +      y          2012        (      y    2    −  2  y  +  1  )  +      z          2012        (      z    2    −  2  z  +  1  )  =  0  ⟹    ⟹      x          2012        (  x  −  1      )    2    +      y          2012        (  y  −  1      )    2    +      z          2012        (  z  −  1      )    2    =  0Since all the sumands are product of even powers, they cannot be negative, so they all are   1. That means that each unknown is either   0 or   1. But only if they all are   1 the original equations hold.Generalize this to   n equations of the same kind and   n unknowns:      {                                                x            1            a                                    +                                      x            2            a                                    +                                      x            3            a                                    +                                    …                          +                                      x            n            a                                    =                          n                                                          x            1                          a              +              1                                                +                                      x            2                          a              +              1                                                +                                      x            3                          a              +              1                                                +                                    …                          +                                      x            n                          a              +              1                                                =                          n                                                          x            1                          a              +              2                                                +                                      x            2                          a              +              2                                                +                                      x            3                          a              +              2                                                +                                    …                          +                                      x            n                          a              +              2                                                =                          n                                                        ⋮                                            ⋮                                            ⋮                                            ⋱                                            ⋮                                            ⋮                                                          x            1                          a              +              n              −              1                                                +                                      x            2                          a              +              n              −              1                                                +                                      x            3                          a              +              n              −              1                                                +                                    …                          +                                      x            n                          a              +              n              −              1                                                =                          n                        With   a even and   n odd.Then we wan define       e    j   as the   L  H  S of the   j-th equation. Then:      ∑          k      =      1        n                      (                    n      k                      )              (  −  1      )          k      +      1            e    k    =      ∑          k      =      1              n            x    k    a    (      x    k    −  1      )          n      −      1        =  0So same as before, each       x    k   is either   1 or   0, but only the   n-tuple   (  1  ,  1  ,  ⋯  ,  1  ) satisfies the equation.If all the RHS are   m  &amp;lt;  n with   m a positive integer, then the solutions are the   n-tuples with   m ones and   n  −  m zeros permutated.

Ilnaus5 · Accepted Answer

While the method already posted is nice, can&#039;t resist this hint - from the first and last equation, by the inequality of power means                              x                      2014                          +                  y                      2014                          +                  z                      2014                              3        2014    ⩾                              x                      2012                          +                  y                      2012                          +                  z                      2012                              3        2012  with equality iff       |    x      |    =      |    y      |    =      |    z      |    =  1. Now using the second equation, the unique answer is obvious.

Find all triplets (x,y,z) that satisfy: x^2012+y^2012+z^2012=3, x^2013+y^2013+z^2013=3, x^2014+y^2014+z^2014=3 Trivial solution: (1,1,1)

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