What happens if no final subtraction is done in Montgomery multiplication?
Theorem 2 in [Dussé et al. 1991] states that, if we skip the final subtraction, then, for $N &lt; R / 4$ and $0 \leq A, B &lt; 2 N$, we have $0 \leq C = \text{MonMul}(A B) &lt; 2 N$, while keeping $C \equiv A B R^{-1} \pmod N$. I think the condition $N &lt; R / 4$ inherently holds in your case e.g. you are using larger $R$ value acutually.