The prime index-ratio divisor-pair conjecture
The prime index-ratio divisor-pair conjecture
Let be a prime number, and let be a -index ratio number, with divisors . Prime index-ratio divisor-pair conjecture. Then
Theorem 9 proves this when , and the source notes that infinitely many examples with larger divisor count satisfy the equalities. The general statement is supported by numerical calculations and remains open.
Progress summary
The conjecture has a small-case proof, but an unverified reader-provided claim gives counterexamples that would disprove it, including one in the supposedly settled case.
The conjecture says that, for these numbers and a prime ratio, the increasing divisors must form consecutive pairs with common multiplier . Mittou’s paper introduces this problem and records the small-divisor theorem, larger examples, and numerical support for the general claim.
Known results
- Theorem 9 proves the pairing for .
- Infinitely many examples with satisfy all the asserted pairings.
- Numerical calculations support the general conjecture.
Posted attempt
An unverified complete-counterexample claim gives , with divisor sums ratio but , contradicting the claimed theorem. It also gives , where but ; neither calculation has independent verification here.
Current status (as of August 2026): The general conjecture has no verified proof or disproof; the posted examples, if correct, would overturn even the stated result.
Sources
Sources & referencesView supporting material
Primary source
Brahim Mittou, “A New Classification of Positive Integers Via New Divisor Functions”, arXiv:2509.08844 (2025).
Solutions 1
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Complete counterexamples, including the natural strengthened hypothesis.
For increasing positive divisors , write
Let . Its complete divisor list is
Consequently
Thus is prime, but the first required divisor-pair identity fails:
The conjecture is therefore false. Since , this also contradicts Theorem 9 in the cited paper under its stated hypothesis .
Even imposing the additional condition does not repair the claim. Take
whose complete divisor list is
Here
so , yet
Hence even the strengthened divisor-pair statement is false.