Proof. We begin by proving (ii). Merca showed that (3) holds if $m \in \{2,3,6\}$ [Reference Merca3, Theorem 3]. Hence, it suffices to show that if $m \notin \{2,3,6\}$ , then there exists some $n \in \mathbb {Z}^+$ such that $n \neq P_5({\kern2pt}j)$ for all $j \in \mathbb {Z}$ and

(5) $$ \begin{align} \sum_{k = -\infty}^\infty \sigma_{\mathrm{odd}}(n-P_5(k)) \not\equiv 0 \pmod m. \end{align} $$

Since $\sigma _{\mathrm {odd}}(n-P_5(k)) = 0$ whenever $n-P_5(k) \le 0$ , the sum in (5) is in fact finite and we easily compute $\sum _k \sigma _{\mathrm {odd}}(3-P_5(k)) = 6$ , where $3 \ne P_5({\kern2pt}j)$ for $j \in \mathbb {Z}$ . Thus, (5) holds unless $6 \equiv 0 \pmod m$ . But this is the case only if $m \in \{2,3,6\}$ .

Next, we prove (iii). Again, Merca proved that (4) holds if $m \in \{2,4\}$ [Reference Merca3, Theorem 4]. Hence, it suffices to show that if $m \notin \{2,4\}$ , then there exists some $n \in \mathbb {Z}^+$ such that $n\neq P_5({\kern2pt}j)$ for all $j \in \mathbb {Z}$ and

(6) $$ \begin{align} \sum_{k = -\infty}^\infty (-1)^{P_3(-k)} \sigma_{\mathrm{odd}}(n-P_5(k)) \not\equiv 0 \pmod m. \end{align} $$

We compute $\sum _k (-1)^{P_3(-k)} \sigma _{\mathrm {odd}}(3-P_5(k)) = 4$ , where $3 \ne P_5({\kern2pt}j)$ for $j \in \mathbb {Z}$ , and so (6) holds unless $4 \equiv 0 \pmod {m}$ . But this is the case only if $m \in \{2,4\}$ .

Finally, we prove (i). Since $\sigma _{\mathrm {odd}}(n)$ is odd if and only if n is a square or twice a square (see [Reference Merca3, page 3]),

(7) $$ \begin{align} \sum_{n=1}^\infty \sigma_{\mathrm{odd}}(n)q^n \equiv \sum_{n=1}^\infty q^{n^2}+\sum_{n=1}^\infty q^{2n^2} \pmod 2. \end{align} $$

The nth coefficient of

$$ \begin{align*} \bigg(\sum_{\ell=1}^\infty \sigma_{\mathrm{odd}}(\ell)q^\ell\bigg)\bigg(\sum_{k=-\infty}^\infty q^{P_m(k)}\bigg) = \sum_{\ell=1}^\infty \sum_{k=-\infty}^\infty \sigma_{\mathrm{odd}}(\ell)q^{\ell+P_m(k)} {=} \sum_{n = 1}^\infty \bigg(\sum_{k=-\infty}^\infty \sigma_{\mathrm{odd}}(n-P_m(k))\bigg)q^n \end{align*} $$

is given by $\sum _{k=-\infty }^\infty \sigma _{\mathrm {odd}}(n-P_m(k))$ . On the other hand, the nth coefficient of

$$ \begin{align*} \bigg(\sum_{\ell=1}^\infty q^{\ell^2}+\sum_{\ell=1}^\infty q^{2\ell^2}\bigg)\bigg(\sum_{k=-\infty}^\infty q^{P_m(k)}\bigg) = \sum_{\substack{\ell \ge 1 \\ k \in \mathbb{Z}}} (q^{\ell^2+P_m(k)}+q^{2\ell^2+P_m(k)}) \end{align*} $$

is given by $a_m(n)+b_m(n)$ , where

$$ \begin{align*} a_m(n) = |A_m(n)| &:= \#\{(\ell,k) \in \mathbb{Z}^+\times \mathbb{Z}:\, \ell^2+P_m(k) = n\}, \\ b_m(n) = |B_m(n)| &:= \#\{(\ell,k) \in \mathbb{Z}^+\times \mathbb{Z}:\, 2\ell^2+P_m(k) = n\}. \end{align*} $$

Thus, due to (7),

$$ \begin{align*} \sum_{k=-\infty}^\infty \sigma_{\mathrm{odd}}(n-P_m(k)) \equiv a_m(n)+b_m(n) \pmod 2. \end{align*} $$

Suppose first that $m \ge 7$ . Then we claim that $P_m(0) = 0, P_m(1) = 1$ and $P_m(k)> 3$ for all $k \notin \{0,1\}$ . From (1), it is clear that $P_m(0) = 0$ and $P_m(1) = 1$ . To see that $P_m(k)> 3$ for all $k \notin \{0,1\}$ , note that since the leading term of $P_m(x)$ is positive and its minimum is at $(m-4)/(2m-4)$ , where $0 < {(m-4)}/{(2m-4)} < 1$ , we have $P_m(k) \ge P_m(2) = m \ge 7$ for $k \ge 2$ and $P_m(k) \ge P_m(-1) = m-3 \ge 4$ for $k \le -1$ .

Now, let $n=3$ . Then the above shows that n is not a generalised m-gonal number for $m \ge 7$ , and so for (2) to hold, we must have $\sum _{k=-\infty }^\infty \sigma _{\mathrm {odd}}(3-P_m(k)) \equiv 0 \pmod 2$ . If $(\ell ,k) \in A_m(3)$ , then $\ell ^2 = 3-P_m(k)$ , so that in particular $\ell ^2 \le 3$ , which forces $\ell = 1$ . But then we must have $P_m(k) = 2$ , which we have seen to be impossible. Hence, $A_m(3)$ is empty and $a_m(3) \equiv 0 \pmod 2$ . On the other hand, if $(\ell ,k) \in B_m(3)$ , we must again have $\ell = 1$ . It follows that $P_m(k) = 1$ , which is the case if and only if $k = 1$ . Hence, $B_m(3) = \{(1,1)\}$ and $b_m(3) \equiv 1\pmod 2$ . We conclude that $\sum _{k=-\infty }^\infty \sigma _{\mathrm {odd}}(3-P_m(k)) \equiv a_m(3)+b_m(3)\equiv 1 \not \equiv 0 \pmod 2$ .

Merca showed that (2) holds for $m \in \{5,6\}$ and, for $m \in \{1,2\}$ , the sum in (2) diverges; hence, it remains to consider $m \in \{3,4\}$ . Suppose first that $m = 3$ and note that $P_3(k) = \tfrac 12(k^2+k)$ . We have $3 = P_3(-3) = P_3(2)$ , so for (2) to hold, we must have $\sum _{k=-\infty }^\infty \sigma _{\mathrm {odd}}(3-P_3(k)) \equiv 3 \equiv 1 \pmod 2$ . If $(\ell ,k) \in A_3(3)$ , then $\ell = 1$ and $P_3(k) = 2$ , which is impossible. Hence, $A_3(3)$ is empty. If $(\ell ,k) \in B_3(3)$ , then $\ell =1$ and $P_3(k) = 1$ , which is the case if and only if $k \in \{-2,1\}$ . It follows that $B_3(3) = \{(1,-2),(1,1)\}$ and $\sum _{k=-\infty }^\infty \sigma _{\mathrm {odd}}(3-P_3(k)) \equiv a_3(3)+b_3(3) \equiv 0 \not \equiv 1 \pmod 2$ .

Finally, suppose that $m = 4$ and note that $P_4(k) = k^2$ . Since $4 = P_4(2)$ , for (2) to hold we must have $\sum _{k=-\infty }^\infty \sigma _{\mathrm {odd}}(4-P_4(k)) \equiv 4 \equiv 0 \pmod 2$ . If $(\ell ,k) \in A_4(4)$ , then either $\ell = 1$ and $P_4(k) = 3$ , which is impossible, or $\ell = 2$ and $P_4(k) = 0$ , which is the case if and only if $k = 0$ . Thus, $A_4(4) = \{(2,0)\}$ . On the other hand, if $(\ell ,k) \in B_4(4)$ , then $\ell =1$ and $P_3(k) = 2$ , which is impossible. It follows that $B_4(4)$ is empty and $\sum _{k=-\infty }^\infty \sigma _{\mathrm {odd}}(4-P_4(k)) \equiv a_4(4)+b_4(4) \equiv 1 \not \equiv 0 \pmod 2$ .