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 valuesa = 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?