Determine all possible real solutions to this system of equations:

A+B+C+D = 5, and:
A*B+B*C+C*D+D*A = 4, and:
A*B*C+B*C*D+C*D*A+D*A*B= 3, and:
A*B*C*D= -1

Prove that there are no others.

Writing the first two equations as (A + C) + (B + D) = 5
                                    and       (A + C)(B + D) = 4
and eliminating B + D gives a quadratic equation in (A + C):

            (A + C)² – 5(A + C) + 4 = 0

with solution      (A + C, B + D) = (1, 4)  or  (4, 1).

First consider A + C = 1, B + D = 4.

Sub. C = 1 – A and D = 4 – B into the 3rd and 4th equations:

  The 3rd equation   AB(C + D) + CD(A + B) = 3
                       
     becomes           AB(5 – A – B) + (1 – A)(4 – B)(A + B) = 3

and simplifies to    (B – 2)² = 4A – 4A² + 1                       (1)

  The 4th equation   AB(1 – A)(4 – B) = -1

    becomes           (B – 2)² = 4 + 1/(A(1 – A))                  (2)

Using (1) and (2):          4A – 4A² + 1 = 4 + 1/(A(1 – A))

which simplifies to          4A⁴ – 8A³ + 7A² – 3A - 1 = 0

Thus                             (A² – A + 1)(4A² – 4A + 1) = 0

The first factor yields no real roots, but the second factor gives
the two real roots:
                        A = (1 + sqrt2)/2 and (1 – sqrt2)/2

For both of these, substitution in (1) gives B = 2 and C and D can
then be found from earlier work to give the two following answers:

            {A, B, C, D} = {(1 + sqrt2)/2,  2,  (1 – sqrt2)/2,  2}
or         {A, B, C, D} = {(1 - sqrt2)/2,  2,  (1 + sqrt2)/2,  2}

Second consideration of  A + C = 4, B + D = 1 gives two further
answers (otherwise found by cycling A->B, B->A, C->D, D->C):

            {A, B, C, D} = {2, (1 + sqrt2)/2, 2, (1 – sqrt2)/2}
            {A, B, C, D} = {2, (1 – sqrt2)/2,  2, (1 + sqrt2)/2}

All possibilities considered and all real roots retained but I don’t
know a simple way of checking that no other real solutions exist.

Posted by Harry on 2016-08-11 16:31:53