Thread: Algebra.. with 8 variables.

1. Algebra.. with 8 variables.

Ok i have these 4 equations:
a^2 + bc + w^2 + xy = (a+w)^2 + (b+x)(c+y)

ca + cd + yw + yz = (c+y)(a+w) + (c+y)(d+z)

ba + bd + xw + xz = (b+x)(a+w) + (b+x)(d+z)

cb + d^2 + yx +z^2 = (c+y)(b+x) + (d+z)^2
From those i have derived:
0 = 2aw + by + xc (1)

0 = 2dz + by + xc (2)

0 = c(w+z) + y(a+d) (3)

0 = b(w+z) + x(a+d) (4)
I'm trying to find what values (integers only) satisfy each of these equations..

so i do (1) - (2)
0 = 2aw - 2dz

2dz = 2aw

dz = aw
(3)/(4)
0 = (c+y) / (b+x)
Thus
c+y = 0

b+x /= 0

dz = aw
Deriving some values:
a = 3, b = 2, c = 4, d = 6
w = 4, x = 1, y = -4, z = 2
None of these satisfy the original equations.
So trying to gather some more rules:
(3)+(4)
0 = c(w+z) + y(a+d) + b(w+z) + x(a+d)

0 = (c+B)(w+z) + (y+x)(a+d)

(c+B)(w+z) = -(y+x)(a+d)

(c+B)(w+z)/(a+d) = -(y+x)
So give random values to a and d
a = 2, d = 4
Derive w and z from this (the equation aw = dz)
w = 6 z = 3

Sub in:
((c+B)(6+3))/(2+4) = -(y+x)
-(y+x) must be an integer so:

Give values to c and b so that:
9(c+B)/(6) = int
b = 1 c = 3

Thus
-(y+x) = 6
c + y = 0 so:
1 + y = 0, so y = -1
Sub in:
-(-1+x) = 6

-1 + x = -6

x = -5
So using the values
a = 2, b = 3, c = 1, d = 4, w = 6, x = -5, y = -1, z = 3
and subbing back into the original 4 equations.....
IS WRONG!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! !!!!!!!
so can anyone see where my error/s are?

2. (3)/(4)

0 = (c+y) / (b+x)
Take a closer look.