MATH.11:5.3 - Change the value arithmetic to find a parity obstruction

The state is an integer n. Allowed steps add 2 or subtract 2. Starting at zero, can the construction reach 1?

A rational-valued linear candidate I(n)=a*n+b changes by 2*a under addition of 2. Requiring zero forces a=0, leaving only constants in this family.

Instead take I(n)=n mod 2, with values 0 and 1. Both +2 and -2 preserve the remainder. The start has remainder 0 and the target remainder 1, proving impossibility.

Here the necessary condition also leads to a construction. For any even target n=2k, use k additions of 2 when k≥0, or -k subtractions of 2 when k<0. Thus the reachable states are exactly the even integers. The invariant and the explicit sequence establish the two directions of that statement.

Adding the steps +1 and -1 destroys this parity invariant: 0 can now move to 1. Every integer becomes reachable by repeated unit steps. The earlier two-step rules and their parity calculation remain correct, but no longer cover all allowed steps.