Solving a Single Congruence Equation

[Han]

In keeping the algorithm fairly simple, I was thinking of the following adjustment. If \(A > B\) then compute \( (q,r) \in \mathbb{Z}^2 \) such that \( A = qB + r\). Then
\[ Ai - B \equiv (qB+r)i - B \equiv B (qi -1) + ri \]
and let \( i \) run from \( -N/2 \) to \( N/2 \) (adjusting for even/odd \( N \) of course).

Similarly, if \( B = qA + r \) then
\[ Ai - B \equiv Ai - (qA+r) \equiv A (i-q) -r \]

We still run into overflow issues, but I think this approach might handle a few more cases than the original approach.

Also, I wonder if the MOD command would work better than division inside FP().