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Cryptography

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    $\begingroup$ As the question stands now, the Second Least Significant Bit can be impossible to compute. Example: $b=2$, $p=19$, $r=4$, solutions include $x=2$ and $x=20$, and these have different Second Least Significant Bit. One fix is to add the constraint $1\le x<p$. Also, we want $b\not\equiv-1\pmod p$, and more generally that $b$ is a generator. $\endgroup$ Commented Nov 5 at 17:20
  • $\begingroup$ I have updated to add the constraint on $x$. b is any possible integer so I am unclear what else is needed to clarify. Please feel free to update. $\endgroup$ Commented Nov 5 at 18:35
  • $\begingroup$ Example of why we need restriction on $b$ [beside $b\ne p-1$ ]: $b=6$, $p=31$, $r=5$. Solution is $x\in\{8,14,20,26\}$ and again the Second Least Significant Bit in undefined. To fix this for good we can restrict to $b$ a generator as I first suggested [equivalently: such that $b^{(p-1)/s}\bmod p\ne1$ for all primes $s$ dividing $p-1$ ] and $1\le x<p$ [or $0\le x<p-1$ ]. Another option is to restrict to $1\le x\le\operatorname{ord}(b)$ [or $0\le x<\operatorname{ord}(b)$ ]. Yet another is to define that $x$ is the smallest positive [or non-negative] solution. $\endgroup$ Commented Nov 5 at 19:27