Shunia's closed-form formula for nontrivial integer roots
Let satisfy , , and suppose there is no such that . Then Shunia's conjecture.
The formula gives an elementary closed form for the th root of when the root is nontrivial, under the stated size restriction on . The supplied text attributes the conjecture to Shunia but provides no evidence that it has been proved or disproved, so its resolution remains open.
References
Primary source
Mihai Prunescu and Joseph M. Shunia, “Elementary closed-forms for non-trivial divisors”, arXiv:2510.26939 (2025).
Progress summary
A paper leaves the conjecture open, but a complete proof was posted in the supplied discussion and has not been independently checked.
Shunia’s conjecture asserts a closed formula for the integer part of an th root when , , , and is not an exact th power. Prunescu and Shunia present it as Conjecture 6.1 in their October 2025 preprint, without proving it.
Posted attempt
A reader-written argument claims a complete proof across the full stated parameter range, using polynomial remainders, roots-of-unity estimates, and an error bound separating the resulting ratio from adjacent integers. The argument has not been independently verified.
Current status (as of August 2026): The conjecture has a posted but unverified complete-proof claim; no independently corroborated proof or counterexample is recorded.
Sources
Solutions 1
ProofThis solution needs a summarySee full solution
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.