Write every nonzero field element as \(2^t\). The congruences \[ 2^{20}=95,\qquad 2^{50}=100,\qquad 2^{100}=1\pmod {101} \] show that 2 has order 100. A set of field elements is therefore a distinct-product basis exactly when its discrete logarithms form a strict sum cover of \(\mathbf Z/100\mathbf Z\).

If the basis has \(k\) elements, its distinct unordered pairs produce at most \(\binom{k}{2}\) products. Covering all 100 nonzero elements requires \(\binom{k}{2}\geq100\), hence \(k\geq15\).

For the upper bound, take the exponent set \[ B=\{0,1,4,10,16,18,23,29,36,50,57,63,68,70,76,82,85,86\}. \] The 153 sums \(b+b'\pmod {100}\), with \(b,b'\in B\) and \(b\ne b'\), contain every residue class. Equivalently, the field set \[ \{1,2,6,14,16,23,33,49,52,53,59,62,74,78,81,88,90,100\} \] has all 100 nonzero field elements among its products of distinct members. Thus the unknown minimum \(m\) satisfies \[ 15\leq m\leq18. \] This improves the candidate record's certified upper bound of 19. It leaves sizes 15, 16, and 17 unresolved.

Haanpää calls a subset \(S\) of a finite abelian group a strict sum cover when every group element is \(a+b\) for distinct \(a,b\in S\). This is exactly the exponent form of the present problem. His exhaustive backtracking computation covers abelian groups through order 90. Fitch and Jamison's earlier census covers cyclic groups through order 54. Neither published range reaches \(\mathbf Z/100\mathbf Z\).

A 2021 discussion on Mathematics Stack Exchange posed \(f_{\rm cyc}(100)\) as a concrete challenge with the same counting lower bound. A comment there establishes \(f_+(100)=18\) under a convention that only requires the nonzero classes. The present artifact independently checks all 100 classes required here.

For an exact test below 18, introduce Boolean variables \(x_i\) for selecting \(i\in\mathbf Z/100\mathbf Z\). For every target \(t\), auxiliary variables encode the unordered pairs \(\{i,j\}\) with \(i+j=t\), and a coverage clause requires at least one such pair. Cardinality clauses impose at most 17 selected residues. Affine symmetry permits the normalization \(x_0=x_1=1\): every cover has a pair whose difference is coprime to 100, since otherwise the parity and modulo-5 projections force the set into one parity class or one residue class modulo 5. Translation and multiplication by the inverse difference send that pair to 0 and 1.

The normalized SAT runs exceeded the research timebox without returning a model or an unsatisfiability certificate. This is computational silence. It supplies no lower bound beyond 15. A complete size-17 model would improve the upper bound, while an independently replayable unsatisfiability certificate would reduce the remaining cases.

The program first certifies that 2 has order 100 by checking the two prime divisors of 100. It then computes the powers indexed by the displayed exponent set. Every unordered pair is evaluated once in the exponent group and once in the field. Both images must equal their full target sets.

The multiplicity distribution of the modular sums is 70 classes represented once, 15 represented twice, 12 represented three times, two represented four times, and one represented nine times. These counts account for all 153 pairs. The canonical output line has SHA-256 digest `90143a42f6b43f9dde66574827f9954b8b90d7b9a43aa5d2d40d14e840c68ff0`.