Let $(\frac{a}{p})$ denote the Legendre symbol, and let $\psi(q)=\sum_{n\geq 0} q^{n(n+1)/2}$.
We define Ramanujan's general theta function $f(a,b)$ for $\mid ab \mid <1$ as $$ f(a,b)=\sum_{n=-\infty}^{\infty}a^{n(n+1)/2}b^{n(n-1)/2}. $$
Let the $p$-dissection of $\psi(q)$:
$$ \psi(q)=\sum_{k= 0}^{(p-3)/2} q^{(k^{2}+k)/2} f(q^{(p^{2}+(2k+1)p)/2}, q^{(p^{2}-(2k+1)p)/2})+q^{(p^{2}-1)/8}\psi(q^{p^{2}}). $$ For any integer $0 \leq k \leq (p-1)/2$, if $$ \dfrac{k^{2}+k}{2} \equiv \dfrac{p^{2}-1}{8} \ (mod \ p), $$ namely, $(2k+1)^{2}\equiv 0 \ (mod \ p)$, then it implies that $k=(p-1)/2$.
The following theorem show the establishement of an infnite family of congruences modulo 3 for the partitions function. I want to include this theorem into my Master's dissertation but i can not fully understand the proof or at least the purpose of some steps. The theorem can be found in : Naika, M.S.M., Shivashankar, C. Arithmetic properties of bipartitions with designated summands. Bol. Soc. Mat. Mex. 24, 37–60 (2018). https://doi.org/10.1007/s40590-016-0140-8 (Theorem 3.4 p.11).
Theorem: For any prime $p \equiv 5 \ (mod \ 6)$, $\alpha \geq 0$, and $n \geq 0$, we have $$ P(6p^{2\alpha +2}+(6i+3p)p^{2\alpha +1}) \equiv 0 \ (mod \ 3), $$ where $i$ is an integer and $1\leq i \leq p-1$.
Proof: We have from previous results that $$ P(6n+3)\equiv \psi(q) \psi(q^{3}) \ (mod \ 3). $$ Define $$ \sum_{n\geq 0}a(n) q^{n} = \psi(q) \psi(q^{3}). $$ Then, $$ P(6n+3)\equiv a(n) \ (mod \ 3). $$ Now consider the congruence equation $$ \dfrac{k^{2}+k}{2}+3\cdot \dfrac{m^{2}+m}{2} \equiv \dfrac{4p^{2}-4}{8}=\dfrac{p^{2}-1}{2} \ (mod \ p), $$ which is equivalent to $$ (2k+1)^{2}+3(2m+1)^{2}\equiv 0 \ (mod \ p), $$ where $0\leq k$, $m\leq (p-1)/2$ and $p$ is a prime such that $(\frac{-3}{p})=-1$. Since $(\frac{-3}{p})=-1$ for $p\equiv 5 \ (mod \ 6)$, the congruence relation $$ (2k+1)^{2}+3(2m+1)^{2}\equiv 0 \ (mod \ p), $$ holds if and only if $k=m=(p-1)/2$.
Therefore, if we substitue the $p$-dissection of $\psi(q)$ into $$ \sum_{n\geq 0}a(n) q^{n} = \psi(q) \psi(q^{3}). $$ and then extracting the terms in which the powers of $q$ are congruent to $(p^{2}-1)/2$ modulo $p$ and then devide by $q^{(p^{2}-1)/2}$, we find that $$ \sum_{n\geq 0}a(pn+(p^{2}-1)/2)q^{pn}=\psi(q^{p^{2}}) \psi(q^{3p^{2}}) $$ which implies that $$ \sum_{n\geq 0}a(p^{2}n+(p^{2}-1)/2)q^{n}=\psi(q) \psi(q^{3}) $$ and for $n\geq 0$, $$ \sum_{n\geq 0}a(p^{2}n+pi+(p^{2}-1)/2)=0, $$ where $i$ is an integer and $1\leq i \leq p-1$. By induction we see that for $n\geq 0$ and $\alpha \geq 0$, $$ \sum_{n\geq 0}a(p^{2\alpha}n+(p^{2\alpha}-1)/2)=a(n). $$
Replacing $n$ by $p^{2}n+pi+(p^{2}-1)/2$ in the above equation and using
$$ \sum_{n\geq 0}a(p^{2}n+pi+(p^{2}-1)/2)=0, $$ we find that for $n\geq 0$ and $\alpha \geq 0$,
$$ \sum_{n\geq 0}a(p^{2\alpha+2}n+p^{2\alpha+1}i+(p^{2\alpha+2}-1)/2)=0. $$ Again, replacing $n$ by $p^{2\alpha+2}n+p^{2\alpha+1}i+(p^{2\alpha+2}-1)/2$ ($i=1,2,\dots, p-1$) in $$ P(6n+3)\equiv \psi(q) \psi(q^{3}) \ (mod \ 3). $$ we arrive at $$ P(6p^{2\alpha +2}+(6i+3p)p^{2\alpha +1}) \equiv 0 \ (mod \ 3). $$
My questions are:
What is the role of Legendre symbol in this theorem? and why we choosed $(\frac{-3}{p})=-1$?.
How $$ \sum_{n\geq 0}a(pn+(p^{2}-1)/2)q^{pn}=\psi(q^{p^{2}}) \psi(q^{3p^{2}}) $$ implies that $$ \sum_{n\geq 0}a(p^{2}n+(p^{2}-1)/2)q^{n}=\psi(q) \psi(q^{3}) $$?.
How we arrived at $$ \sum_{n\geq 0}a(p^{2}n+pi+(p^{2}-1)/2)=0. $$
How we arrived at $$ \sum_{n\geq 0}a(p^{2\alpha}n+(p^{2\alpha}-1)/2)=a(n). $$