The $N=16n+6$ divisor-sum congruence conjecture

Less than 1 year old · traced to 2026

Let $N=16n+6$ for some nonnegative integer $n$ . Define $\a0\sigma_0(m)$ to be the number of positive divisors of $m$ . The $N=16n+6$ congruence conjecture.

\sum_{1\leq k<\sqrt{N}}\sigma_0(N-k^2)\equiv0\pmod4.

The authors report computational evidence for this congruence for small values of $N$ and conjecture that it extends the theorem, whose currently established conditions on $N$ are more restrictive. The general case remains open.

References

Primary source

Sittinon Jirattikansakul, Teeradej Kittipassorn, Kraiwich Kongsiri, Nitipon Moonwichit and Kirati Sriamorn, “Congruences via Partitions with Exactly Two Part Sizes”, arXiv:2604.25394 (2026).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A 2026 paper proves related narrower cases, while an unverified posted argument claims the conjecture follows from a broader result covering the entire progression divisible by eight after adding six.

The conjecture asks whether the divisor sum is always divisible by four when the input has the form $16n+6$ . The 2026 paper by Sittinon Jirattikansakul, Teeradej Kittipassorn, Kraiwich Kongsiri, Nitipon Moonwichit, and Kirati Sriamorn proves several other arithmetic progressions but does not claim this case.

Known results

Jirattikansakul, Kittipassorn, Kongsiri, Moonwichit, and Sriamorn (2026) prove the congruence for $N=16n+14$ , $36n+30$ , $72n+42$ , $196n+70$ , and $252n+114$ .

Posted attempt

An attempted proof claims the congruence for every $N\equiv6\pmod8$ , thereby including the conjectured progression and the known $16n+14$ progression. It is presented as a complete proof, but it has not been independently verified.

Current status (as of August 2026): The published paper settles several narrower progressions, while the $N=16n+6$ conjecture has only an unverified complete-proof claim and therefore remains mathematically unconfirmed.

Sources

arxiv.org

ProofThis solution needs a summarySee full solutionHide full solution

In fact the conjecture holds on the larger progression $N\equiv6\pmod8$ , simultaneously covering both $N=16n+6$ and $N=16n+14$ .

For such $N$ , define

D(N) = \#\{(e,a)\in\mathbb Z_{>0}^2: e\text{ even},\ a\text{ odd},\ e^2+2a^2=N\}.

We prove separately that the odd- and even-index contributions both equal $2D(N)\pmod4$ .

Let $\chi_4$ be the nontrivial character modulo $4$ . If $m\equiv5\pmod8$ , then $m$ is not a square and its divisors pair without fixed points as $\{d,m/d\}$ . Because $m\equiv1\pmod4$ , both divisors in each pair have the same $\chi_4$ -value. Thus each pair contributes $2$ to the divisor count and either $2$ or $-2$ to the character sum. Therefore

d(m)\equiv\sum_{d\mid m}\chi_4(d) =\frac{r_2(m)}4\pmod4,

where

r_2(m)=\#\{(u,v)\in\mathbb Z^2:u^2+v^2=m\}

and the equality is Jacobi's two-square formula. For $m\equiv5\pmod8$ , exactly one coordinate is odd and the other is nonzero even. Hence $r_2(m)/4$ counts ordered positive representations.

Write

S_{\mathrm{odd}} = \sum_{\substack{1\le j<\sqrt N\j\ \mathrm{odd}}} d(N-j^2).

For odd $j$ , $N-j^2\equiv5\pmod8$ . The preceding identity gives

S_{\mathrm{odd}}\equiv2C(N)\pmod4,

where

C(N) = \#\{(j,a,e)\in\mathbb Z_{>0}^3: j,a\text{ odd},\ e\text{ even},\ j^2+a^2+e^2=N\}.

The involution $(j,a,e)\mapsto(a,j,e)$ pairs all off-diagonal triples. Its fixed points satisfy $j=a$ and are counted by $D(N)$ . Consequently

S_{\mathrm{odd}}\equiv2D(N)\pmod4.

Now let $j>0$ be even. Since $N-j^2\equiv2$ or $6\pmod8$ , write $N-j^2=2w$ with $w$ odd. Then

d(N-j^2)=2d(w) \equiv \begin{cases} 2\pmod4,&w\text{ is a square},\\ 0\pmod4,&\text{otherwise}. \end{cases}

The square cases are exactly $j^2+2a^2=N$ with $j$ even and $a$ positive odd. Therefore

S_{\mathrm{even}}\equiv2D(N)\pmod4.

Adding the two contributions yields

\boxed{ \sum_{1\le j<\sqrt N}d(N-j^2) \equiv4D(N)\equiv0\pmod4 \qquad(N\equiv6\pmod8). }

Taking $N=16n+6$ proves Conjecture 7 for every $n\ge0$ ; the same theorem also recovers the previously established progression $N=16n+14$ .

Source: “Congruences via Partitions with Exactly Two Part Sizes,” arXiv:2604.25394, Conjecture 7.