Yes, but the field of order 4 is not the same thing as the integers modulo 4. The field of order 4 has the four elements 0,1,x,x+1, where x satisfies the equation x^2 + x + 1 = 0 and 2=0. For more details see https://en.wikipedia.org/wiki/Finite_field#Field_with_four_elements
The math jargon for a set where you can implement a 4 function calculator (\(+,-,*,/\)) is a field. Fields can have a finite number of elements, and such fields always have a prime power number of elements, aka \(p^k\) where \(p\) is prime and \(k\) is a natural number.
Many results are not true if the underlying field has exactly 2 elements. The reason why:
\(x = -x\) \(x + x = -x + x\) \(2x = 0\)
The final formula is taken to mean that \(x\) is zero. But there’s another interpretation.
What if 2 = 0?
In a field with 2 elements, this is true. Moreover, \(x = -x\) is a tautology AKA always true AKA worthless. \(-1 \mod 2 = 1\) and \(0 \mod 2 = 0\).
This is only when the underlying field has 2 elements. A 2 element set is exceptional because negation doesn’t actually do anything.