I've been working through the following problem and I'm really stuck
Starting with the following three equations:
$$ a= (129A \oplus C)\mod 256 \\ b= (129B \oplus A) \mod 256\\ c= (129C \oplus B) \mod 256 $$ I xor'd both sides of the equation and translated this to this (modulus removed for simplicity): $$ C = 129*A \oplus a \\ A = 129*B \oplus b \\ B = 129*C \oplus c $$ Following this, I used substitution to find the following: $$ A = 129*B \oplus b \\ \mathsf{\text{Substituting for }}B \\ A=129(129C \oplus c) \oplus b\\ \mathsf{\text{Now substituting for }}C \\ A=129(129(129A \oplus a) \oplus c) \oplus b\\ $$
Which then simplifies to: $$ A=(129^3A \oplus 129^2 a \oplus 129 c \oplus b) \mod 256 $$
When I substitute for actual values, the above expression works.
Then
$$ (A\oplus 129^3A )\mod 256= (\oplus 129^2 a \oplus 129 c \oplus b) \mod 256 $$
either equals 0 or 128.
Test values: $[A,B,C] = [223, 12,59]$ and $[a,b,c] = [100, 211, 183]$
EDIT A is an integer between 0 and 255, so I beieve that $A\oplus129^3A$ simplifies to $A\oplus129A$ becuse $\mod256$ would wipe out the unnecessary significant bits
Question
1. Is there any way to isolate $A$ from $A\oplus(129^3A)$? Or, possibly $A\oplus129A$?
2. Have I made some fundamental error that I'm not catching?