AB^4+CD^4 = 2021

AB^3+CD^3 = 485

AB^2+CD^2 = 101

AB+CD = 5

If you were to multiply one of the equations throughout by (B + D) you will get an equation involving some of the other terms on the left-hand side, along with BD and (B+D), for instance:

AB^3+CD^3 = 485, multiplied by (B+D):

1) AB^4+CD^4 + BD(AB^2+CD^2) = 485(B+D)

AB^2+CD^2 = 101, multiplied by (B+D):

2) AB^3+CD^3 + BD(AB+CD) = 101(B+D)

AB+CD = 5, multiplied by (B+D):

3) AB^2+CD^2 + BD(A+C)= 5(B+D)

In each of the equations let (B+D)=E and BD=F, and replace each of the ABCD terms with their numerical value, if known:

1) 2021 + 101F = 485E

2) 485 + 5F = 101E

3) 101 + (A+C)F = 5E

Multiplying 1) by 101 and 2) by 485 and subtracting one from the other gives an equation involving large numbers and F, but neatly divides to give F = 4. Plugging that back in to, say 2), gives a value of E = 5. Plugging both of these values into 3) gives a value of -19 for (A+C).

To find B and D we need to find a pair of numbers whose sum is 5 and whose product is 4. These must be 1 and 4. Let us assume for a moment that B is 4 and D is 1.

So we know that A+C is -19 and 4A+C is 5. Therefore A = 8 and C = -27.

(If we were to have assumed that B is 1 and D is 4, we would get A = -27 and C = 8, so the same four numbers but in a different order.)

Since we are asked for ABCD we get the same answer either way: -864.