Is there any relation between $(q^4\ ;q^4)_\infty$ and $(q\ ;q)^4_\infty$ ?
I have established a claim that, Let, $\{a_n\}$ be a sequence of integers then the coefficients of $q^{7n-1}, n\ge 0$ in $$\displaystyle M(q):= \frac{\sum_{n = 0}^{\infty} a_n q^{n^2}}{(q;q)_\infty^4}$$ is divisible by $7$.
Proof to the claim:
Given, $$\displaystyle M(q):= \frac{\sum_{n = 0}^{\infty} a_n q^{n^2}}{(q;q)_\infty^4}$$
then, $$\displaystyle \frac{(q;q)_\infty^3 \sum_{n=0}^\infty a_nq^{n^2}}{(q;q)_\infty^7}\equiv \frac{(q;q)_\infty^3 \sum_{n=0}^\infty a_nq^{n^2}}{(q^7;q^7)_\infty} \pmod{7}$$
We can see that it is sufficient to examine the coefficient of $q^{7n-1}$ in $$\displaystyle (q;q)_\infty^3 \sum_{n=0}^\infty a_n q^{n^2} = \sum_{n=0}^\infty (-1)^j(2j+1)q^{j(j+1)\over 2}\sum_{n=0}^\infty a_nq^{n^2}$$ i.e. we have to examine the coefficient of $q^{(7n-1)}$ in $$\displaystyle \sum_{n=0}^\infty \sum_{n=0}^\infty a_n (-1)^j (2j+1)q^{\frac{j(j+1)}{2} +n^2}$$
Now, we need those terms for which $$\frac{j(j+1)}{2} + n^2 = 7n-1 \\ \implies \frac{j(j+1)}{2} + n^2 + 1 \equiv 0 \pmod{7} \\ \implies 4j^2 + 4j = n^2 + 1 \equiv 0 \pmod{7} \\ \implies (2j+1)^2 + n^2 \equiv 0 \pmod{7} $$
Now, $(2j+1) \equiv 0,1,2,4 \pmod{7}$ and $n^2 \equiv 0,1,2,4 \pmod{7}$.
So, we must have $(2j+1) \equiv n \equiv 0 \pmod{7}$,
The coefficients of $q^{7n-1}$ in $\displaystyle (q;q)_\infty^3 \sum_{n=0}^\infty a_n q^{n^2}$ is $(-1)^j(2j+1)a_n$ and we have $(2j+1)\equiv 0 \pmod{7}$. So, the coefficients of $q^{(7n-1)}$ in $M(q)$ is divisible by $7$.
Now, using this I have to prove that $P(7n+5) \equiv 0 \pmod{7}$. How, should I proceed to obtain the required result ?
The proof I am seeking is similar to the proof in Page 33 of the book "Number Theory in the spirit of Ramanujan" for $P(5n+4) \equiv 0 \pmod{5}$