Consider the expression $A(x)$ = $B(x)\cdot x + R(x)$, where $A(x)$, $B(x)$, $R(x)$ are polynomials and degree of $R$ is less than degree of $B$, so $R$ is the remainder of long dividing $A$ by $B$. The start point is the pair $(A, B)$, in the first step you will change this pair to ($B$, $A$ mod $B$), then in the second step to ($A$ mod $B$, $B$ mod ($A$ mod $B$), and so on (like in gcd) until the second element in the pair is $0$.
suppose the starting pair $A$, $B$ take $N$ steps to reach the point where the second element in the pair is $0$. the pair $A\%2$ and $B\%2$ also take $N$ steps, but what is exactly the reason that they take the same number of steps?
Note that $A\%2$ and $B\%2$ mean applying mod 2 to all coefficients in $A$ and $B$.
EDIT: A constraint exists where the leading element in $A$ and $B$ (the element with highest power) should have a coefficient of 1, and Degree of $A$ should be higher than Degree of $B$.
EDIT2: Also there is a constraint that the last pair (in the last step) be (1,0), like fibonacci polynomials.