Shunia's closed-form formula for nontrivial integer roots

From papers

Let $n,m\in\mathbb{N}$ satisfy $n>2$ , $\lfloor\log_2(n)\rfloor+1\geq m>1$ , and suppose there is no $k\in\mathbb{N}$ such that $k^m=n$ . Then Shunia's conjecture.

\left\lfloor\sqrt[m]{n}\right\rfloor=\left\lfloor\frac{(n^{2nm}+1)^{2nm+1}\bmod(n^{2nm^2}-n)}{(n^{2nm}+1)^{2nm}\bmod(n^{2nm^2}-n)}-1\right\rfloor.

The formula gives an elementary closed form for the $m$ th root of $n$ when the root is nontrivial, under the stated size restriction on $m$ . The supplied text attributes the conjecture to Shunia but provides no evidence that it has been proved or disproved, so its resolution remains open.

Progress summary

Open

A 2025 paper repeats the proposed formula but does not prove it, and no disproof was found.

Shunia’s conjecture asserts that a modular-exponential expression equals the integer part of the nontrivial $m$ th root of $n$ under the stated bounds. It is attributed to Shunia, but no proof or disproof is documented.

October 2025 preprint

The preprint Elementary closed-forms for non-trivial divisors reproduces the statement as “Conjecture 6.1 (Shunia)” and says that, if proved, it would simplify its main result. It neither proves nor refutes this particular formula; no retrieved source reports a verified resolution.

Current status (as of August 2026): The formula remains an open conjecture; no proof, counterexample, or independent verification is recorded in the retrieved sources.

Sources

arxiv.org

Sources & referencesView supporting material

Primary source

Mihai Prunescu and Joseph M. Shunia, “Elementary closed-forms for non-trivial divisors”, arXiv:2510.26939 (2025).

1 week ago

The later source denotes its radicand by N and its exponent by m. Set a=N and n=m in the following argument; its domain and modular expression then coincide exactly with the stated conjecture.

Let a>2 and 1<n≤⌊log₂a⌋+1, with a not an nth power. Put α=a^{1/n}, k=2an, b=a^k, M=b^n−a, q=1+α. We prove ⌊α⌋=⌊(((b+1)^{k+1} mod M)/((b+1)^k mod M))−1⌋.

Define P_j(X)=(1+X)^j mod (X^n−a)=Σ_{r=0}^{n−1}c_{j,r}X^r, c_{j,r}=Σ_{h≥0}C(j,r+hn)a^h. These coefficients are nonnegative and Σ_r c_{j,r}≤P_j(α)=q^j. In fact q^{k+1}<b: if a≥4, then q≤1+√a≤3a/4, k≥4a, and (4/3)^{4a}≥1+4a/3>a; the sole remaining case a=3,n=2 follows from 11^13<3^12·4^13 and √3<7/4. Hence for j=k,k+1, 0<P_j(b)≤b^{n−1}(b−1)<b^n−a=M. Since b^n≡a mod M, both modular residues are therefore exactly P_j(b).

Let ζ=e^{2πi/n}. Roots-of-unity filtering gives c_{k,r}=q^k/(nα^r)·(1+ε_r), ε_r=Σ_{s=1}^{n−1}ζ^{−rs}((1+αζ^s)/(1+α))^k. Set ρ²=1−[4α/(1+α)²]sin²(π/n), E=(n−1)ρ^k. Then |ε_r|≤E. Multiplication by 1+X modulo X^n−a gives c_{k+1,0}=c_{k,0}+ac_{k,n−1} and c_{k+1,r}=c_{k,r}+c_{k,r−1} for r≥1. With w_r=(b/α)^r, P_{k+1}(b)/P_k(b)−1 =α[Σ_rw_r(1+ε_{r−1 mod n})]/[Σ_rw_r(1+ε_r)]. Consequently, whenever E<1, |P_{k+1}(b)/P_k(b)−1−α|≤2αE/(1−E).

We now establish uniformly E<1/[4αn(1+α)^{n−1}]. (*) For n=2, ρ=(α−1)/(α+1), k=4α², and log((α+1)/(α−1))≥2/α, so E≤e^{−8α}. Since α≥√3>3/2, monotonicity gives 8α(1+α)E<30e^{−12}<1.

For n≥3, a≥2^{n−1}, hence α≥2^{(n−1)/n}. Using sin(π/n)≥3√3/(2n), (1+α)²<3α², and log(1−t)≤−t gives ρ^k<exp(−9α^{n−1}/n). It suffices to show H_n(x)=4n(n−1)x(1+x)^{n−1}exp(−9x^{n−1}/n)<1 for x≥2^{(n−1)/n}. Differentiation shows H_n is decreasing there. For n=3 its endpoint is bounded by 432e^{−7}<1. For n≥4 the endpoint is bounded by K_n=8n(n−1)3^{n−1}exp(−9·2^{n−2}/n). Here K_4=2592e^{−9}<1 and K_{n+1}/K_n=[3(n+1)/(n−1)]exp(−9·2^{n−2}(n−1)/(n(n+1)))<5e^{−5}<1. This proves (*).

Finally let t=⌊α⌋. Since a is not an nth power, both a−t^n and (t+1)^n−a are positive integers. Factoring differences of powers gives min(α−t,t+1−α)≥δ=1/[n(1+α)^{n−1}]. By (*), E<δ/(4α)<1/4, so |P_{k+1}(b)/P_k(b)−1−α|<2δ/3<δ. Thus P_{k+1}(b)/P_k(b)−1 lies in the same open unit interval (t,t+1) as α. Taking floors and substituting the exact modular residues proves the formula throughout the full conjectured parameter range.

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Shivam Patel · 21 hours ago