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

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.