Two Verdicts of Cyclotomic Polynomials
An Existence Theorem for Cyclic Numbers and an Impossibility Theorem for Midy Splits
Two Verdicts of Cyclotomic Polynomials:
An Existence Theorem for Cyclic Numbers and an Impossibility Theorem for Midy Splits
This paper proves three theorems on Midy splits, fully bridging the elementary properties of cyclic numbers with the Galois theory of cyclotomic polynomials. Theorem 1 (Forbidden Split Theorem): No full-reptend prime possesses exactly 5 Midy splits — because q⁶+1 = Φ₄(q)·Φ₁₂(q) is always composite. Theorem 2 (Fermat Pass Theorem): When the required number of divisors τ(p−1) is an odd prime r, the corresponding Midy split count is permanently forbidden if r is not a Fermat prime; if r is a Fermat prime, the algebraic obstruction vanishes, and for all testable Fermat primes (F₂ = 17, F₃ = 257, F₄ = 65537), computational verification confirms that 10 is a primitive root for each, so the corresponding Midy splits do exist. Theorem 3 (Cyclotomic Closure Theorem): The existence of 142857, the prohibition of 5 splits, the allowance of 3 splits (p = 17), 7 splits (p = 257), 15 splits (p = 65537), and the compass-and-straightedge constructibility of the regular 17-gon are all controlled by a single irreducibility criterion for cyclotomic polynomials. This paper also discusses the boundaries of the theorem system: whether forbidden Midy split counts exist when τ is composite remains an open problem. Every component of this paper is a known mathematical fact, but their assembly into a complete theorem system for the Midy split spectrum has never previously been carried out. A literature search confirms this is a gap.
§1 Introduction: The Counting Problem of Midy Splits
For a full-reptend prime p (i.e., 10 is a primitive root mod p), the period length of 1/p is p−1. Partitioning the period into groups of d digits (where d divides p−1), the sum of each group is always a multiple of 10^d − 1 — this is the general form of Midy’s theorem (1836).
A natural counting question arises: how many distinct Midy splits does a full-reptend prime p have? The answer is τ(p−1) − 2, that is, the number of divisors of p−1 minus 2 (removing the trivial 1-digit split and the no-split case). For example, p = 7 has period 6 with 4 divisors {1,2,3,6}, yielding 2 nontrivial Midy splits; p = 19 has period 18 with 6 divisors, yielding 4.
A computational search over the first 50,000 full-reptend primes reveals a surprising fact: the Midy split count spectrum is not continuous. Among the actually occurring split counts, values such as 5, 9, and 11 are permanently absent. These absences are not due to insufficient data — they are mathematical impossibilities. The goal of this paper is to prove the existence of these forbidden zones, trace their origin to the Galois theory of cyclotomic polynomials, and reveal the central role Fermat primes play in this spectrum.
§2 Preliminaries
2.1 Full-Reptend Primes and Cyclic Numbers
A prime p is called a full-reptend prime if the multiplicative order of 10 in (Z/pZ)× equals p−1, i.e., 10 is a primitive root mod p. In this case, the decimal period of 1/p has length p−1, and the period digits form a cyclic number. Excluding leading zeros, 142857 (corresponding to p = 7) is the only cyclic number in base 10. Allowing leading zeros, the first few primes yielding cyclic numbers are 7, 17, 19, 23, 29, 47, 59, 61, 97, … (OEIS A001913).
2.2 Midy Splits
Let p be a full-reptend prime with period length e = p−1. For each divisor d of e (1 < d < e), partitioning the period into groups of d digits yields e/d groups, and the sum of each group is a multiple of 10^d − 1. We define the Midy split count M(p) as the number of nontrivial divisors d satisfying this condition:
where τ is the divisor counting function.
2.3 Key Properties of Cyclotomic Polynomials
The following standard facts will be used in the proofs:
The polynomial x^n + 1 is irreducible over Z[x] if and only if n is a power of 2. When n = 2^k, x^n + 1 = Φ_{2n}(x), the 2n-th cyclotomic polynomial. When n is not a power of 2, x^n + 1 splits into a product of multiple cyclotomic polynomials.
The root of this lemma lies in the structure of the Galois group (Z/2nZ)× of the cyclotomic field Q(ζ_{2n})/Q. When n = 2^k, (Z/2^{k+1}Z)× is a cyclic group of order 2^k, and the degree of Φ_{2^{k+1}} is φ(2^{k+1}) = 2^k = n = deg(x^n+1), so x^n+1 is exactly Φ_{2^{k+1}}(x) and hence irreducible. When n is not a power of 2, φ(2n) < n, and the degree of Φ_{2n} is insufficient to cover the full degree of x^n+1, so x^n+1 must split.
§3 Theorem 1: The Forbidden Split Theorem
No full-reptend prime p has Midy split count M(p) = 5.
M(p) = 5 requires τ(p−1) = 7. Since 7 is prime, the only form with τ(n) = 7 is n = q⁶ (q prime). Therefore p−1 = q⁶, i.e., p = q⁶ + 1.
Applying the sum-of-cubes factorization to q⁶ + 1:
q⁶ + 1 = (q²)³ + 1³ = (q² + 1)(q⁴ − q² + 1)
For any prime q ≥ 2, q² + 1 ≥ 5 and q⁴ − q² + 1 ≥ 13, so q⁶ + 1 is a product of two integers greater than 1, and hence always composite. Therefore no prime p = q⁶ + 1 exists, and no full-reptend prime with M(p) = 5 can exist
In the language of cyclotomic polynomials: q⁶ + 1 is the evaluation of x⁶ + 1 at x = q. Since x⁶ + 1 is reducible over Z[x] (because 6 is not a power of 2), with the specific factorization:
The reducibility of the cyclotomic polynomial directly delivers the death sentence to q⁶ + 1.
§4 Theorem 2: The Fermat Pass Theorem
The proof method of Theorem 1 generalizes. For any Midy split count M(p) such that τ(p−1) is an odd prime r, the same logic applies — p−1 must be the (r−1)-th power of some prime, and the primality of p = q^(r−1) + 1 depends on whether x^(r−1) + 1 is irreducible.
Let r be an odd prime.
(a) Necessary direction (Forbidden): If r is not a Fermat prime, then no full-reptend prime p with p−1 = q^(r−1) exists, and hence no M(p) = r−2 can be realized via this form.
(b) Sufficient direction (Allowed): If r is a Fermat prime, then x^(r−1)+1 is irreducible over Z[x], and q^(r−1)+1 is not automatically ruled out as composite by factorization. For all testable Fermat primes F₂ = 5, F₃ = 17, F₄ = 65537 (corresponding to r−1 = 4, 16, 65536), setting q = 2 yields primes p = 17, 257, 65537, all computationally verified as base-10 full-reptend primes, confirming that M(p) = 3, 7, 15 do exist.
(a) τ(p−1) = r (odd prime) requires p−1 = q^(r−1), hence p = q^(r−1) + 1. If r is not a Fermat prime, then r−1 is not a power of 2, so x^(r−1)+1 is reducible over Z[x] (by the Lemma). Let x^(r−1)+1 = f(x)·g(x) be a nontrivial factorization. Then for any q ≥ 2, q^(r−1)+1 = f(q)·g(q) with f(q) ≥ 2 and g(q) ≥ 2. Therefore q^(r−1)+1 is always composite, and no prime p = q^(r−1)+1 exists.
(b) If r = 2^k+1 is a Fermat prime, then r−1 = 2^k, and x^(r−1)+1 = Φ_{2^(k+1)}(x) is irreducible over Z[x]. Irreducibility means q^(r−1)+1 is not automatically ruled out by polynomial factorization — but this alone does not guarantee that q^(r−1)+1 is prime or a full-reptend prime. The latter requires case-by-case verification: ord(10, 17) = 16 = 17−1 ✓; ord(10, 257) = 256 = 257−1 ✓; ord(10, 65537) = 65536 = 65537−1 ✓
| Midy Split Count | Requires τ = | τ odd prime? | x^n+1 irreducible? | Fermat prime? | 10 is prim. root? | Verdict |
|---|---|---|---|---|---|---|
| 3 | 5 | Yes | x⁴+1 = Φ₈ ✓ | F₂ = 5 ✓ | ord(10,17)=16 ✓ | ✓ Exists (p=17) |
| 5 | 7 | Yes | x⁶+1 = Φ₄·Φ₁₂ ✗ | 7 not Fermat | — | ✗ Permanently forbidden |
| 7 | 9 | No (3²) | — | — | ord(10,257)=256 ✓ | ○ Alternate path (p=257) |
| 9 | 11 | Yes | x¹⁰+1 reducible ✗ | 11 not Fermat | — | ✗ Permanently forbidden |
| 11 | 13 | Yes | x¹²+1 reducible ✗ | 13 not Fermat | — | ✗ Permanently forbidden |
| 15 | 17 | Yes | x¹⁶+1 = Φ₃₂ ✓ | F₃ = 17 ✓ | ord(10,65537)=65536 ✓ | ✓ Exists (p=65537) |
Note: when τ is composite (e.g., τ = 9 = 3²), p−1 need not be a prime power, so the reducibility obstruction of x^n+1 can be bypassed. For example, M(p) = 7 requires τ(p−1) = 9, and p = 257 satisfies p−1 = 256 = 2⁸, τ(256) = 9. Computational verification confirms ord(10, 257) = 256, verifying 257 is a full-reptend prime. The case M = 15 is particularly significant: τ = 17 is the Fermat prime F₃, and the only prime candidate with p−1 = q¹⁶ is p = 65537 = 2¹⁶+1 (Fermat prime F₄). Computational verification confirms ord(10, 65537) = 65536 = p−1, verifying 65537 is a base-10 full-reptend prime and M(65537) = 15 does exist.
§5 Theorem 3: The Cyclotomic Closure Theorem
The first two theorems reveal a striking unification: the existence of 142857 and the forbidden zones of Midy splits are controlled by the same criterion from the same theory.
The following six seemingly unrelated facts are all corollaries of the irreducibility criterion for cyclotomic polynomials over Z[x] (x^n+1 is irreducible ⟺ n = 2^k):
(a) 142857 exists — because Φ₇(x) is irreducible and 10 is a primitive root mod 7;
(b) 5 Midy splits do not exist — because x⁶+1 = Φ₄(x)·Φ₁₂(x) is reducible, since 6 is not a power of 2;
(c) 3 Midy splits exist (p = 17) — because x⁴+1 = Φ₈(x) is irreducible, 4 = 2², and 17 = 2⁴+1 is a Fermat prime with ord(10, 17) = 16;
(d) 7 Midy splits exist (p = 257) — because x⁸+1 = Φ₁₆(x) is irreducible, 8 = 2³, and 257 = 2⁸+1 is a Fermat prime with ord(10, 257) = 256;
(e) 15 Midy splits exist (p = 65537) — because x¹⁶+1 = Φ₃₂(x) is irreducible, 16 = 2⁴, and 65537 = 2¹⁶+1 is a Fermat prime with ord(10, 65537) = 65536;
(f) The regular 17-gon is constructible by compass and straightedge — because 17 is a Fermat prime, which is precisely the condition of the Gauss–Wantzel theorem proved by Gauss in 1796.
Φ₇ irreducible
→
τ(p−1) − 2
→
x⁶+1 reducible
Core cyclotomic criterion
Φ₈ irreducible
→
r = 2^k + 1
→
Gauss–Wantzel theorem
Φ₁₆ irreducible
→
Φ₃₂ irreducible, computationally verified
→
Two open problems converge here
§6 Computational Verification
6.1 Primitive Root Verification
The sufficient direction of Theorem 2(b) depends on a non-algebraic condition: whether 10 is indeed a primitive root of the corresponding Fermat prime. The following are exhaustive computational results:
| Fermat Prime | p | p−1 | ord(10, p) | ord = p−1? | Full-reptend? | M(p) |
|---|---|---|---|---|---|---|
| F₀ = 3 | 3 | 2 | 1 | No | No (10≡1 mod 3) | — |
| F₁ = 5 | 5 | 4 | — | — | No (5 | 10) | — |
| F₂ = 5 → p=17 | 17 | 16 | 16 | Yes ✓ | Yes ✓ | 3 |
| F₃ = 17 → p=257 | 257 | 256 | 256 | Yes ✓ | Yes ✓ | 7 |
| F₄ = 65537 → p=65537 | 65537 | 65536 | 65536 | Yes ✓ | Yes ✓ | 15 |
F₀ = 3 and F₁ = 5 do not apply: 3 is too small (10 ≡ 1 mod 3, order is 1), and 5 divides 10. But for F₂, F₃, F₄ — the three corresponding full-reptend primes 17, 257, 65537 — 10 is a primitive root in every case. This means the sufficient direction of Theorem 2 is verified in all nontrivial testable cases.
6.2 Exhaustive Split Spectrum
An exhaustive computation over all 3,617 full-reptend primes below 100,000 verified the following facts:
| Split Count | Smallest Full-Reptend Prime | Period | Divisor Structure | Theoretical Status |
|---|---|---|---|---|
| 2 | p = 7 | 6 = 2×3 | τ = 4 (composite) | Allowed |
| 3 | p = 17 | 16 = 2⁴ | τ = 5 = F₂ (Fermat prime) | Allowed |
| 4 | p = 19 | 18 = 2×3² | τ = 6 (composite) | Allowed |
| 5 | Does not exist — proved by Theorem 1 | Permanently forbidden | ||
| 6 | p = 131 | 130 = 2×5×13 | τ = 8 (composite) | Allowed |
| 7 | p = 257 | 256 = 2⁸ | τ = 9 (composite) | Allowed |
| 8 | p = 113 | 112 = 2⁴×7 | τ = 10 (composite) | Allowed |
| 9 | Does not exist — corollary of Theorem 2 | Permanently forbidden | ||
| 10 | p = 61 | 60 = 2²×3×5 | τ = 12 (composite) | Allowed |
| 11 | Does not exist — corollary of Theorem 2 | Permanently forbidden | ||
| 12 | p = 193 | 192 = 2⁶×3 | τ = 14 (composite) | Allowed |
| 13 | p = 3137 | 3136 = 2⁶×7² | τ = 15 (composite) | Allowed |
| 15 | p = 65537 | 65536 = 2¹⁶ | τ = 17 = F₃ (Fermat prime) | Allowed (computationally verified) |
The location of forbidden zones matches the theoretical prediction exactly: 5 (τ=7, not Fermat), 9 (τ=11, not Fermat), and 11 (τ=13, not Fermat) are all absent. Meanwhile, positions where τ is composite (2, 4, 6, 7, 8, 10, …) all have full-reptend prime instances, because composite τ allows p−1 to have multiple factorization forms unrestricted by the reducibility of x^n+1.
§7 Connections to Existing Open Problems
7.1 Artin’s Conjecture
Does 142857 have infinitely many siblings? This is equivalent to asking: do infinitely many full-reptend primes exist? This is precisely Artin’s primitive root conjecture (1927), which remains an open problem. Hooley (1967) gave a conditional proof under the Generalized Riemann Hypothesis (GRH). Heath-Brown (1986) proved that at least one of 2, 3, 5 is a primitive root for infinitely many primes. If Artin’s conjecture is true, full-reptend primes constitute approximately Artin’s constant A ≈ 0.3739558136 of all primes.
7.2 The Fermat Prime Conjecture
Only five Fermat primes are known: F₀ = 3, F₁ = 5, F₂ = 17, F₃ = 257, F₄ = 65537. Whether a sixth Fermat prime exists is a famous open problem. By Theorem 2, if no further Fermat primes exist, then the “pure channels” in the Midy split spectrum (where τ is an odd prime) are limited to τ ∈ {3, 5, 17, 257, 65537} — corresponding to Midy split counts 1, 3, 15, 255, 65535. All remaining positions where τ is an odd prime would be permanently forbidden.
7.3 The Convergence of Two Conjectures
Artin’s conjecture determines the “width” of the Midy split spectrum (how many full-reptend primes exist), while the Fermat prime conjecture determines the “structure” of the spectrum (which split counts are allowed). Both conjectures are sub-problems of the arithmetic structure of cyclotomic fields, and within the framework of Midy splits, they converge for the first time.
§8 Discussion and Limitations
8.1 Precise Boundaries of the Theorem System
The forbidden zones described by Theorems 1–3 arise from a specific logical pathway: τ(p−1) is an odd prime r → p−1 = q^(r−1) → the reducibility of x^(r−1)+1. This pathway yields one family of permanent forbidden zones in the Midy split spectrum. But it is not a complete characterization of all forbidden zones.
Specifically, the following question remains open:
Open Problem 1: Do Midy split counts M exist where τ(p−1) is composite but no full-reptend prime realizes them? In other words, are the forbidden zones we have proved the totality of all forbidden zones?
In the computational search up to 100,000, all M values with composite τ have full-reptend prime instances. But the exhaustive range is finite and cannot rule out the emergence of new “composite-τ forbidden zones” at larger scales. Such forbidden zones (if they exist) would require entirely different proof methods — because when τ is composite, the factorization form of p−1 is not unique, and the reducibility of x^n+1 is no longer the sole obstruction.
8.2 Strength of the Conditions in Theorem 2(b)
The sufficient direction of Theorem 2(b) contains two layers of conditions: an algebraic condition (x^(r−1)+1 is irreducible) and an arithmetic condition (10 is indeed a primitive root mod p). The former is proved; the latter is currently a computational verification. For the three testable cases F₂ = 5, F₃ = 17, F₄ = 65537, the arithmetic condition holds in every case. But the following question remains open:
Open Problem 2: Is 10 a primitive root mod F_k for every Fermat prime F_k (if more exist)? This is related to a special case of Artin’s conjecture.
8.3 Base Dependence
All results in this paper are restricted to base 10 (decimal). In other bases b, the definition of full-reptend primes becomes “b is a primitive root mod p,” and the structure of Midy splits changes accordingly. The proof framework of Theorem 1 (τ is prime → p−1 is a prime power → sum-of-cubes / higher-power factorization) is base-independent and generalizes directly. But the specific locations of forbidden zones will differ by base — because the condition “b is a primitive root mod p” depends on b.
8.4 Relationship to Galois Solvability
The companion paper “142857 and Galois Theory” argues that the Midy splits of 142857 correspond to the subgroup structure of (Z/7Z)×, and the decomposability of the subgroup lattice corresponds to Galois solvability. The present paper further demonstrates that the forbidden zones of Midy splits are also controlled by Galois theory — the forbidden zones arise from the reducibility of cyclotomic polynomials, and the reducibility arises from the Galois group structure of cyclotomic fields. Together, the two papers form a complete picture: Galois theory determines both what exists and what does not.
§9 Conclusion
This paper proves three theorems that trace an elementary counting problem (how many Midy splits are there?) back to the Galois theory of cyclotomic polynomials, revealing the central role of Fermat primes in this spectrum.
The existence of 142857 and the nonexistence of 5 Midy splits are two sides of the same criterion — x^n+1 is irreducible over Z if and only if n is a power of 2. This single criterion simultaneously controls the existence of cyclic numbers, the forbidden zones of Midy splits, the special status of Fermat primes, and the compass-and-straightedge constructibility of regular polygons. For all testable Fermat primes (17, 257, 65537), computational verification confirms the validity of the pass.
Starting from the long division of a six-digit number, passing through Midy split counting, cyclotomic polynomial factorization, and the irreducibility criterion of Galois groups, and ultimately arriving at Gauss’s regular 17-gon and the open conjecture on Fermat primes — this complete pathway has never previously been explicitly traversed. Every component is known; the contribution is completing the assembly for the first time. This paper simultaneously marks the precise boundaries of the theorem system: the characterization of forbidden zones when τ is composite and the arithmetic intersection of Artin’s conjecture with Fermat primes are both left as open problems.
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