Hello I am trying to prove this congruence:
$$P(7n+5)\equiv 0 \pmod{7}$$
In order to do that I have done the next thing:
We have that
$\displaystyle\sum_{n\geq0}\;P(n)q^{n}=\frac{1}{(q;q)_{\infty}}$
we multiply by $q^{2}$ then
\begin{eqnarray} \displaystyle\sum_{n\geq0}\;P(n)q^{n+2}&=&\frac{q^{2}}{(q;q)_{\infty}}\\ &=& \frac{q^{2}((q;q)^{3}_{\infty})^{2}}{(q;q)_{\infty}^{7}} \end{eqnarray}
Therefore the coefficient of $q^{7n+7}$ in the LHS is $P(7n+5)$ therefore we have to check that the coefficient of $q^{7n+7}$ of $ \frac{q^{2}((q;q)^{3}_{\infty})^{2}}{(q;q)_{\infty}^{7}}$ is $\equiv 0(mod \; 7)$.
Now we have that $(q;q)^{3}_{\infty}=\displaystyle\sum_{n\geq0} (-1)^{n}(2n+1)q^{\frac{n(n+1)}{2}}$, therefore
\begin{eqnarray} q^{2}((q;q)^{3}_{\infty})^{2}&=&(q(q;q)^{3}_{\infty})^{2}\\ &=& \displaystyle \sum_{n,m \geq0} (-1)^{n}(2n+1)(2m+1)q^{\frac{n(n+1)}{2}+\frac{m(m+1)}{2}+2} \end{eqnarray}
Now we will check when $\frac{n(n+1)}{2}+\frac{m(m+1)}{2}+2$ is a mutiply of $7$
Note that $$(2n+1)^{2}+(2m+1)^{2}=8\left(\frac{n(n+1)}{2}+\frac{m(m+1)}{2}+2\right)-14$$
Then $\frac{n(n+1)}{2}+\frac{m(m+1)}{2}+2\equiv 0(mod\;7)$ if and only if $(2n+1)^{2}+(2m+1)^{2}\equiv0(mod \;7)$ only if $(2n+1)^{2}\equiv0(mod \;7)$ and $(2m+1)^{2}\equiv0(mod \;7)$ Then $2n+1\equiv0(mod \;7)$ and $2m+1\equiv0(mod \;7)$.
Therefore the coefficient of $q^{7n+7}$ in $q^{2}((q;q)^{3}_{\infty})^{2}$ is is multiple of 7.
I do not know if that is correct this idea and also I do not know how to do it for $\frac{1}{(q;q)^{7}_{\infty}}$. I think I can use $$\frac{1}{(1-q)^{7}}\equiv \frac{1}{1-q^{7}}(mod\;7)$$ so that I can have $$\frac{1}{(q;q)^{7}_{\infty}}\equiv\frac{1}{(q^{7};q^{7})_{\infty}}(mod\; 7 )$$
But I do not know how to procede with this. I woud appreciate any hint you can give me.
Thank you for your time!
I think the bit you're missing is $$\frac{1}{1 - z} = 1 + z + z^2 + \cdots$$ Therefore $$\frac{1}{(q^7;q^7)_\infty} = \prod_{k=1}^\infty \sum_{j=0}^\infty q^{7jk}$$ which clearly only has non-zero coefficients for powers of seven.
Thus the coefficient $[q^{7n+7}]\frac{q^{2}((q;q)^{3}_{\infty})^{2}}{(q^{7};q^{7})_{\infty}}$ is some integer-weighted sum* of coefficients $[q^{7i}]q^{2}((q;q)^{3}_{\infty})^{2}$.
* It is, of course, quite easy to be more precise than this, but unnecessary for the argument. On the other hand, maybe it's a nicer argument to say that $\frac{1}{(q^7;q^7)_\infty}$ is the generating function for the partition numbers inflated by a factor of seven, so that $$[q^{7n+7}]\frac{q^{2}((q;q)^{3}_{\infty})^{2}}{(q^{7};q^{7})_{\infty}} = [q^{7n+7}](q^{2}((q;q)^{3}_{\infty})^{2})\sum_{i=0}^\infty P(i)q^{7i} = \sum_{i=0}^{n+1} P(i)[q^{7(n-i+1)}](q^{2}((q;q)^{3}_{\infty})^{2})$$