Is it immediately apparent that the solution to the system of equations, (1) x 1 2 = x 2 + 2 x 2 2 = x 3 + 2 x 3 2 = x 4 + 2 ⋮ x n 2 = x 1 + 2 can be given by the roots of unity? Specifically, (2a) x = y k 2 + 1 y k where the y k are, (2b) y k = exp ⁡ ( 2 π i k 2 n − 1 ) , k = 0 … 2 n − 1 − 1 y k = exp ⁡ ( 2 π i k 2 n + 1 ) , k = 1 … 2 n − 1 Example. Let n = 4. Then ( 1 ) is equivalent to, (3) x = ( ( ( x 2 − 2 ) 2 − 2 ) 2 − 2 ) 2 − 2 Expanded out, ( 3 ) is a 2 4 = 16-deg polynomial and its 16 roots are given by ( 2 ) where, y k = exp ⁡ ( 2 π i k 15 ) , k = 0 … 7 y k = exp ⁡ ( 2 π i k 17 ) , k = 1 … 8Ramanujan considered the system ( 1 ) for n = 3 , 4 in the general case and also as nested radicals. For x = ( ( ( x 2 − a ) 2 − a ) 2 − a ) 2 − a, see this related post. (Interestingly, n = 5 in the general case is no longer completely solvable in radicals.)Question:I observed ( 2 ) empirically. How do we prove from first principles that this is indeed the solution?

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Is it immediately apparent that the solution to the system of equations,                    (1)                                                                        x                1                2                                                                  =                              x                2                            +              2                                                                          x                2                2                                                                  =                              x                3                            +              2                                                                          x                3                2                                                                  =                              x                4                            +              2                                                                                    ⋮                                                                          x                n                2                                                                  =                              x                1                            +              2                                          can be given by the roots of unity? Specifically,                    (2a)                    x        =                                            y              k              2                        +            1                                y            k                              where the       y    k   are,                    (2b)                                                                        y                k                                                                  =              exp              ⁡                                            (                                                              2                  π                  i                  k                                                                      2                    n                                    −                  1                                                                          )                            ,                            k              =              0              …                              2                                  n                  −                  1                                            −              1                                                                          y                k                                                                  =              exp              ⁡                                            (                                                              2                  π                  i                  k                                                                      2                    n                                    +                  1                                                                          )                            ,                            k              =              1              …                              2                                  n                  −                  1                                                                        Example. Let   n  =  4. Then   (  1  ) is equivalent to,                    (3)                    x        =        (        (        (                  x          2                −        2                  )          2                −        2                  )          2                −        2                  )          2                −        2            Expanded out,   (  3  ) is a       2    4    =  16-deg polynomial and its 16 roots are given by   (  2  ) where,      y    k    =  exp  ⁡        (              2      π      i      k        15          )    ,    k  =  0  …  7      y    k    =  exp  ⁡        (              2      π      i      k        17          )    ,    k  =  1  …  8Ramanujan considered the system   (  1  ) for   n  =  3  ,  4 in the general case and also as nested radicals. For   x  =  (  (  (      x    2    −  a      )    2    −  a      )    2    −  a      )    2    −  a, see this related post. (Interestingly,   n  =  5 in the general case is no longer completely solvable in radicals.)Question:I observed   (  2  ) empirically. How do we prove from first principles that this is indeed the solution?

tradirasi · Accepted Answer

Let   f  (  x  )  =      x    2    −  2. Then  f  (  2  cos  ⁡  u  )  =  4      cos    2    ⁡  u  −  2  =  2  (  2      cos    2    ⁡  u  −  1  )  =  2  cos  ⁡  2  uso      f    n    (  2  cos  ⁡  u  )  =  2  cos  ⁡      2    n    uThen       f    n    (  x  )  =  x becomes  cos  ⁡  u  =  cos  ⁡      2    n    ufor which solutions are given by  (      2    n    −  1  )  u  =  2  k  π  ,  k  =  0  ,  ±  1  ,  ±  2  ,  …

Is it immediately apparent that the solution to the system of equations, <mtable displaystyle="t

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