Find the real numbers a,b,c,d,e in [−2,2] that simultaneously satisfy the following relations: a + b + c + d + e = 0 a 3 + b 3 + c 3 + d 3 + e 3 = 0 a 5 + b 5 + c 5 + d 5 + e 5 = 10

Question

Find the real numbers a,b,c,d,e in [−2,2] that simultaneously satisfy the following relations:  a  +  b  +  c  +  d  +  e  =  0      a    3    +      b    3    +      c    3    +      d    3    +      e    3    =  0      a    5    +      b    5    +      c    5    +      d    5    +      e    5    =  10

Charlee Gentry · Accepted Answer

The unknowns   a  ,  b  ,  c  ,  d  ,  e are to be real and in the interval   [  −  2  ,  2  ]. This screams for the substitution   a  =  2  cos  ⁡      ϕ    1  ,   b  =  2  cos  ⁡      ϕ    2  ,   …  ,  e  =  2  cos  ⁡      ϕ    5   with some unknown angles       ϕ    j    ,  j  =  1  ,  2  ,  3  ,  4  ,  5 to be made. Let&#039;s use the equations   2  cos  ⁡      ϕ    j    =      e          i              ϕ        j              +      e          −      i              ϕ        j            ,   j  =  1  ,  2  ,  3  ,  4  ,  5. Now  0  =  a  +  b  +  c  +  d  +  e  =      ∑          j      =      1        5    (      e          i              ϕ        j              +      e          −      i              ϕ        j              )  ,Using this in the second equation gives  0  =      a    3    +      b    3    +      c    3    +      d    3    +      e    3    =      ∑          j      =      1        5    (      e          3      i              ϕ        j              +  3      e          i              ϕ        j              +  3      e          −      i              ϕ        j              +      e          −      3      i              ϕ        j              )  =      ∑          j      =      1        5    (      e          3      i              ϕ        j              +      e          −      3      i              ϕ        j              )  .Using both of these in the last equation gives                    10        =                  a          5                +                  b          5                +                  c          5                +                  d          5                +                  e          5                                    =                  ∑                      j            =            1                    5                (                  e                      5            i                          ϕ              j                                      +        5                  e                      3            i                          ϕ              j                                      +        10                  e                      i                          ϕ              j                                      +        10                  e                      −            i                          ϕ              j                                      +        5                  e                      −            3            i                          ϕ              j                                      +                  e                      −            5            i                          ϕ              j                                      )                                          =                  ∑                      j            =            1                    5                (                  e                      5            i                          ϕ              j                                      +                  e                      −            5            i                          ϕ              j                                      )        =                  ∑                      j            =            1                    5                (        2        cos        ⁡        5                  ϕ          j                )        .            This is equivalent to      ∑          j      =      1        5    cos  ⁡  5      ϕ    j    =  5.When we know that the sum of five cosines is equal to five, certain deductions can be made.This shows that there are 5 possible values for all the five unknowns, namely   2  cos  ⁡  (  2  k  π  /  5  ) with   k  =  0  ,  1  ,  2  ,  3  ,  4 (well, cosine is an even function, so there are only three!). We get a solution by using each value of   k exactly once, because then the first two equations are satisfied (use familiar identities involving roots of unity)

sebadillab0 · Accepted Answer

We know that the sum of the 5th roots of unity is 0, i.e. that      ∑          k      =      0        4        e          i      2      π      k          /      5        =  0.What happens if we consider the powers? Turns out that      ∑          k      =      0        4    (      e          i      2      π      k          /      5            )    3    =  0too (to see this, note that   x  ⟼      x    3   is an automorphism since the order of the group is 5), and      ∑          k      =      0        4    (      e          i      2      π      k          /      5            )    5    =      ∑          k      =      0        4    1  =  5.with this knowledge it is simple; scale all the variables by       2          1          /      5      . Take   {      2          1          /      5            e          i      2      π      k          /      5            }          k      =      0        4   for   a  ,  b  ,  c  ,  d  ,  e.

Find the real numbers a,b,c,d,e in [−2,2] that simultaneously satisfy the following relations:

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