While doing some training in optimisation I was hit by the system of polynomial equations (1) t a 6 b + b d 2 = 2 a 3 (2) t b 6 c + c a 2 = 2 b 3 (3) t c 6 d + d b 2 = 2 c 3 (4) t d 6 a + a c 2 = 2 d 3 (5) a 5 + b 5 + c 5 + d 5 = 4 where a , b , c , d , t are real variables being strictly positive.- The system is cyclic in a , b , c , d- The remaining variable t gets easily isolated: Multiply ( 1 ) through ( 4 ) with the variables not present in the t-term, then sum up and use ( 5 ) which yields (6) t ⋅ 4 a b c d = a 3 c d + b 3 d a + c 3 a b + d 3 b c . - There's the solution with all variables = 1.Are there more solutions?

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While doing some training in optimisation I was hit by the system of polynomial equations                    (1)                    t                  a          6                b        +        b                  d          2                                    =                    2                  a          3                                    (2)                    t                  b          6                c        +        c                  a          2                                    =                    2                  b          3                                    (3)                    t                  c          6                d        +        d                  b          2                                    =                    2                  c          3                                    (4)                    t                  d          6                a        +        a                  c          2                                    =                    2                  d          3                                    (5)                              a          5                +                  b          5                +                  c          5                +                  d          5                                    =                    4            where     a  ,  b  ,  c  ,  d  ,  t   are real variables being strictly positive.- The system is cyclic in     a  ,  b  ,  c  ,  d- The remaining variable     t   gets easily isolated: Multiply   (  1  ) through   (  4  ) with the variables not present in the t-term, then sum up and use   (  5  ) which yields                    (6)                    t        ⋅        4        a        b        c        d                =                          a          3                c        d        +                  b          3                d        a        +                  c          3                a        b        +                  d          3                b        c                .            - There&#039;s the solution with all variables   =  1.Are there more solutions?

Arcatuert3u · Accepted Answer

YES, amazingly there is another real and positive solution to the system                    t                  a          6                b        +        b                  d          2                                    =        2                  a          3                                    t                  b          6                c        +        c                  a          2                                    =        2                  b          3                                    t                  c          6                d        +        d                  b          2                                    =        2                  c          3                                    t                  d          6                a        +        a                  c          2                                    =        2                  d          3                                              a          5                +                  b          5                +                  c          5                +                  d          5                                    =        4            On the off-chance there was, I used Mathematica&#039;s Resultant function to reduce this to a single equation in one unknown. The extra solution is  a  =  (  2      x    1        )          1          /      5        =  0.748744  …    b  =  (  2      y    2        )          1          /      5        =  1.24441  …    c  =  (  2      x    2        )          1          /      5        =  0.904281  …    d  =  (  2      y    1        )          1          /      5        =  0.706404  …    t  =            2              a        3            −      b              d        2                            a        6            b        =  0.996756  …    a  b  c  d  =  0.595186  ⋯  &amp;lt;  1    a  b  c  d  t  =  0.593255  ⋯  &amp;lt;  1where       x    1    ,      x    2    ≈  0.117662  ,  0.302334 are the two real roots of the   14-deg,            16      −      1152      y      +      24688              y        2            −      80400              y        3            −      796648              y        4            −      709040              y        5            +      950456              y        6            +      152880              y        7            +      2136              y        8            −      100380              y        9            +      15168              y              10            −      19628              y              11            +      3908              y              12            +      1274              y              13            +      343              y              14            =      0      The   a  ,  b  ,  c  ,  d also satisfy the constraint,        (        a    c                  )        3    =      d    b  while       x    1    ,      x    2   satisfy the simple   7th-deg,      z    7    −  5      z    6    +  7      z    5    −      z    4    −      z    3    −      z    2    −  z  −  1  =  0as       z    =                  1        2                        (                      x        1                    x        2              +                  x        2                    x        1                        )        .

While doing some training in optimisation I was hit by the system of polynomial equations <mtab

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