I know you cannot find $b$ if you are given $B$ and $G$, where $B = [b]G$,

but can you find $G$ given $b$ and $B$?

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    $\begingroup$ If you know the order of $G$, call it $q$, you can compute $b'=b^{-1}\bmod q$ and then $b'\cdot B$ should give you $G$ (if I'm doing my math right, else someone will correct me in another comment / answer, I'm too tired right now for a full answer / guaranteed info) $\endgroup$ – SEJPM Oct 23 '19 at 22:38

$G = [b^{-1} \bmod q]B$ where $q$ is the order of the group generated by $G$, assuming $\gcd(b, q) = 1$.

| improve this answer | |
  • $\begingroup$ How do you find b^-1? $\endgroup$ – user40176 Oct 24 '19 at 14:42
  • $\begingroup$ Use the extended Euclidean algorithm to compute the Bézout coefficients $s$ and $t$ so that $s b + t q = \gcd(b, q)$; then $s$ is your answer. Or, use Fermat's little theorem and compute $b^{q - 2} \bmod q$. $\endgroup$ – Squeamish Ossifrage Oct 24 '19 at 15:48

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