XOR using mathematical operators

Question

How can I implement XOR using basic mathematical operators like +,-,*,/

Update: Actually, I need to track change in two matrix having Boolean values. T

Answer 1

(a − b)²

This works because:

(a − b)² = a * (a − b) + b * (b − a)

Since multiplication in ℤ₂ is conjuction (&), and 1 - a is negation (!), the above formula is equivalent to XOR for a, b ∈ {0, 1}:

(a & !b) | (b & !a)

See the comment below by Pascal Cuoq explaining why this cannot be a linear equation.