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Some context: I was thinking about the feasibility of using SAT solvers to prove primality, especially of Mersenne primes, by showing that there exists no Boolean sequence $d_1,d_2, ..., d_{b'}$ that can represent a divisor of the prime (i.e. UNSAT).

Given a Boolean list $d_1,d_2, ..., d_{b'}$, where $d$ is the base-2 representation of a base-10 positive integers $D$, does there exist a Boolean predicate that evaluates to True if and only if $2^b-1 \equiv 0 ($mod D)$?

(Assume $1 < D < N$, and hence, $b' < b$. Also assume $b$ is prime.)

Any help is greatly appreciated.

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  • $\begingroup$ Isn't it just the case that any boolean-valued function of $b+b'$ boolean variables can be represented using Boolean algebra operations? $\endgroup$ – Daniel Schepler Feb 13 at 0:46
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You could use a circuit which performs a modular reduction as explained in this lecture note (chapter 2.6.4 on page 32). The modular reduction computes the remainder of the division $N/D$. Use a digital comparator circuit to check for zero remainder. When the remainder vanishes to zero, the predicate becomes true.

From the cited lecture note: enter image description here

Note that this might not fully answer your question as it assumes $D$ to be a constant rather than a variable.

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  • $\begingroup$ Close. However, can this be directly evaluated with Boolean logic? I want to implement the predicate directly (i.e. a sequence of logical operations). This is an interesting idea, though. $\endgroup$ – Baaing Cow Feb 12 at 22:52
  • $\begingroup$ Yes. An alternative would be to calculate the remainder $R$ as $R = N - D * (N/D)$. But that would require a divider, a subtractor and a multiplier. $\endgroup$ – Axel Kemper Feb 12 at 22:55
  • $\begingroup$ Subtraction and multiplication of Boolean arrays can be done as if it was in base-2. (subtraction is trivial XOR and NAND for carrying, multiplication can be expanded into addition, i.e. XOR and AND) This question concerns division. $\endgroup$ – Baaing Cow Feb 12 at 23:12
  • $\begingroup$ Divider circuits are actually described in the above-mentioned book Basic Arithmetic Circuits. My point ist that modular reduction is less complex than the classic remainder calculation. $\endgroup$ – Axel Kemper Feb 13 at 7:54

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