When i know the value "c", How do I find factored numbers "a" and "b" : (a*b) == c (mod N) where, N is bitcoin order

When i know the value "c", How do I find factored numbers "a" and "b" : (a*b) == c (mod N) where, N is bitcoin order.

• It's unclear what "factored numbers" are, and the constraints on $a$ and $b$. What about for example $a=1$ and $b=c$? Or any $a\in[1,c)$ and $b=a^{-1}c\bmod N$? Note: the order of secp256k1 is $N=$0xfffffffffffffffffffffffffffffffebaaedce6af48a03bbfd25e8cd0364141 (just below $2^{256}$), and that matches the example given.
• If the question asks how to find some $a$ and $b$ of known factorization and near 256-bit such that $a\,b\equiv c\pmod N$, that's easy, since most 256-bit integers are easy to factor: pick a random $a\in[1,N)$, compute $b=a^{-1}c\bmod N$ (that's b=pow(a,-1,N)*c%N in python), and factor $a$ and $b$ (or try another $a$ in the unlikely case that turns out to be hard). But I don't see the point.
• Because $N$ is prime, any integer $a$ such that $0 < a < N$ has a corresponding value $b$ such that $ab \equiv c \pmod{N}$. You can literally just choose one. There is no way to know what $a$ and $b$ values were used by someone to compute $c$ because any nonzero choice of $a$ could have been used. As a simple example, you could choose $a=1$ in which case $b=c$. fgrieu's comment shows how to compute $b$ given $a$. Mar 28 at 18:34