Conjecture 7. For any odd prime $p$, \begin{equation} \sum_{1\leq n \leq (p-1)p}a(n) \equiv \begin{cases} p \pmod{p^2} &\mbox{if $p \equiv 1 \pmod 4$ }\\ 0 \pmod{p^2} &\mbox{if $p \equiv 3 \pmod 4$ } \end{cases}. \end{equation}\begin{equation} \sum_{n=1}^{p(p-1)}a(n) \equiv \begin{cases} p \pmod{p^2} &\mbox{if $p \equiv 1 \pmod 4$ }\\ 0 \pmod{p^2} &\mbox{if $p \equiv 3 \pmod 4$ } \end{cases}. \end{equation} It is true for $3 \leq p \leq 19$.
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