The sign and congruence conjecture for tangent permanents

About 5 years old · traced to

Let tnt_n and tn′t_n' be the quantities defined earlier in the paper, and let (⋅p)\left(\frac{\cdot}{p}\right) denote the Legendre symbol. The tangent permanent conjecture.**

(i) For every odd composite integer n>1n>1,

tn≡0(modn).t_n\equiv0\pmod n.

(ii) If pp is an odd prime, then

(2p)tp<0\left(\frac2p\right)t_p<0

and

(−1p)tp′<0.\left(\frac{-1}p\right)t_p'<0.

The claim is motivated by the preceding theorem and remark on these tangent quantities; the supplied text gives no resolution.

References

Primary source

Zhi-Wei Sun, “Arithmetic properties of some permanents”, arXiv:2108.07723 (2022).

Progress summary

Refreshed
Claimed solved

A 2026 paper claims the conjecture is settled, but an unverified calculation posted on August 25 gives apparent counterexamples, so its status is disputed.

Sun's Conjecture 4.7 asserts that the tangent permanent vanishes modulo every odd composite integer and that two Legendre-symbol-adjusted prime values are negative. The 2021 paper established integrality and a prime congruence for one quantity, but explicitly left these assertions unresolved.

May 2026 claimed confirmation

A paper dated May 2026 claims to confirm a list of Sun's determinant and permanent conjectures including Conjecture 4.7. The retrieved description does not display the tangent formulas or supply enough proof detail to verify the claimed congruences and signs.

Community submission (unverified; August 25, 2026)

A submitted exact computation argues that the sign assertion for the cotangent permanent fails: it reports positive T29=84,643,985,981,440T_{29}=84{,}643{,}985{,}981{,}440 and negative T43=−7,815,543,081,584,471,417,421,824T_{43}=-7{,}815{,}543{,}081{,}584{,}471{,}417{,}421{,}824, opposite to the predicted signs. This conflicts with the claimed confirmation but has not been checked.

Current status (as of August 2026): A May 2026 paper claims to settle the conjecture, while an August 25 submission alleges sign counterexamples; neither claim is independently verified, so the conjecture is not settled.

Sources

Solutions 1

CounterexampleThis solution needs a summarySee full solutionHide full solution

Let pp be an odd prime, put m=(p−1)/2m=(p-1)/2, and define

Tp=p per⁡[cot⁡(πjkp)]1≤j,k≤m.(1)T_p=\sqrt p\,\operatorname{per}\left[\cot\left(\frac{\pi jk}{p}\right)\right]_{1\le j,k\le m}. \tag{1}

Sun's Theorem 1.7 proves that Tp∈ZT_p\in\mathbb Z and Tp≡1(modp)T_p\equiv1\pmod p. Conjecture 4.7(ii) further predicts

(−1p)Tp<0.(2)\left(\frac{-1}{p}\right)T_p<0. \tag{2}

Both possible congruence classes of pp contradict this prediction:

T29=84,643,985,981,440>0,T43=−7,815,543,081,584,471,417,421,824<0.(3)T_{29}=84{,}643{,}985{,}981{,}440>0, \qquad T_{43}=-7{,}815{,}543{,}081{,}584{,}471{,}417{,}421{,}824<0. \tag{3}

Indeed, 29≡1(mod4)29\equiv1\pmod4, whereas 43≡3(mod4)43\equiv3\pmod4, so both numbers in (3) have the opposite sign to (2).

Here is an entirely exact certificate requiring no numerical approximations. Fix a prime q≡1(mod4p)q\equiv1\pmod{4p}, choose an element w∈Fq×w\in\mathbb F_q^\times of order 4p4p, and set

ζ=w4,ι=wp,G=∑a=0p−1ζa2,g={G,p≡1(mod4),G/ι,p≡3(mod4).(4)\zeta=w^4,\qquad \iota=w^p,\qquad G=\sum_{a=0}^{p-1}\zeta^{a^2},\qquad g=\begin{cases}G,&p\equiv1\pmod4,\\G/\iota,&p\equiv3\pmod4.\end{cases} \tag{4}

The quadratic Gauss-sum evaluation gives g2=pg^2=p; more precisely, gg is the image of the positive real square root appearing in (1) under the cyclotomic reduction. Since

cot⁡(πrp)=i e2πir/p+1e2πir/p−1,(5)\cot\left(\frac{\pi r}{p}\right) =i\,\frac{e^{2\pi i r/p}+1}{e^{2\pi i r/p}-1}, \tag{5}

the image of the cotangent matrix over Fq\mathbb F_q has entries

cjk=ι ζjk+1ζjk−1.(6)c_{jk}=\iota\,\frac{\zeta^{jk}+1}{\zeta^{jk}-1}. \tag{6}

For subsets S⊆{1,…,m}S\subseteq\{1,\ldots,m\}, define recursively

D∅=1,DS=∑k∈Sc∣S∣,kDS∖{k}.(7)D_{\varnothing}=1,\qquad D_S=\sum_{k\in S}c_{|S|,k}D_{S\setminus\{k\}}. \tag{7}

Induction on ∣S∣|S| shows that DSD_S sums exactly the injections from the first ∣S∣|S| rows onto SS. Therefore

Tp≡gD{1,…,m}(modq).(8)T_p\equiv gD_{\{1,\ldots,m\}}\pmod q. \tag{8}

For p=29p=29, formula (7) gives the following three exact residues:

qwT29 mod q1,000,002,361725,178,391786,139,3171,000,003,057194,657,493727,227,7891,000,004,449514,956,177609,404,733(9)\begin{array}{c|r|r} q&w&T_{29}\bmod q\\ \hline 1{,}000{,}002{,}361&725{,}178{,}391&786{,}139{,}317\\ 1{,}000{,}003{,}057&194{,}657{,}493&727{,}227{,}789\\ 1{,}000{,}004{,}449&514{,}956{,}177&609{,}404{,}733 \end{array} \tag{9}

These residues reconstruct 84,643,985,981,44084{,}643{,}985{,}981{,}440 modulo

Q29=1,000,009,867,031,322,291,111,000,073.(10)Q_{29}=1{,}000{,}009{,}867{,}031{,}322{,}291{,}111{,}000{,}073. \tag{10}

To certify uniqueness, write Hm=∑a=1m1/aH_m=\sum_{a=1}^m1/a. The chord bound for sine gives

∣cot⁡(πrp)∣≤p2min⁡(r,p−r).(11)\left|\cot\left(\frac{\pi r}{p}\right)\right| \le\frac{p}{2\min(r,p-r)}. \tag{11}

For every nonzero row index jj, the values min⁡(jk mod p,p−(jk mod p))\min(jk\bmod p,p-(jk\bmod p)), as 1≤k≤m1\le k\le m, are a permutation of 1,…,m1,\ldots,m. Bounding the permanent by the product of its absolute row sums and using p<p\sqrt p<p, we obtain

∣Tp∣<p(p2Hm)m.(12)|T_p|<p\left(\frac p2H_m\right)^m. \tag{12}

At p=29p=29, this yields

∣T29∣<29(33,980,257720,720)14<7,777,485,176,976,639,097,830,653<Q292.(13)|T_{29}|<29\left(\frac{33{,}980{,}257}{720{,}720}\right)^{14} <7{,}777{,}485{,}176{,}976{,}639{,}097{,}830{,}653 <\frac{Q_{29}}2. \tag{13}

Consequently, the Chinese remainder theorem proves the first equality of (3) as an identity of integers.

The same argument for p=43p=43 gives

qT43 mod q1,000,000,433580,932,2151,000,003,013789,445,7101,000,003,357893,159,5441,000,004,3892,428,9471,000,008,173547,386,028(14)\begin{array}{c|r} q&T_{43}\bmod q\\ \hline 1{,}000{,}000{,}433&580{,}932{,}215\\ 1{,}000{,}003{,}013&789{,}445{,}710\\ 1{,}000{,}003{,}357&893{,}159{,}544\\ 1{,}000{,}004{,}389&2{,}428{,}947\\ 1{,}000{,}008{,}173&547{,}386{,}028 \end{array} \tag{14}

Here 43H21/2=810,896,279/10,346,33643H_{21}/2=810{,}896{,}279/10{,}346{,}336, and (12) gives

∣T43∣<257,763,709,733,568,910,737,647,180,249,968,428,462,101<12∏q in (14)q.(15)|T_{43}|<257{,}763{,}709{,}733{,}568{,}910{,}737{,}647{,}180{,}249{,}968{,}428{,}462{,}101 <\frac12\prod_{q\text{ in }(14)}q. \tag{15}

Thus (14) uniquely determines the second integer in (3). Both values also satisfy the independently known congruence Tp≡1(modp)T_p\equiv1\pmod p. The tangent-permanent sign assertion and the odd-composite divisibility assertion in the other parts of Conjecture 4.7 are not addressed here.

Source. Z.-W. Sun, Arithmetic properties of some permanents, Theorem 1.7 and Conjecture 4.7. The previously reported counterexample to the adjacent Conjecture 4.6 concerns sine/cosecant permanents and does not concern the cotangent permanents in (1).