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


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!

  • $\begingroup$ To be clear: you're satisfied with your proof that $$[q^{7n+7}]q^{2}((q;q)^{3}_{\infty})^{2} \equiv 0 \pmod 7$$ and you want to show that $$[q^{7n+7}]\frac{q^{2}((q;q)^{3}_{\infty})^{2}}{(q^{7};q^{7})_{\infty}} \equiv 0 \pmod 7$$? $\endgroup$ – Peter Taylor Dec 5 '18 at 16:33
  • $\begingroup$ Of course but I do not know how to finish it! $\endgroup$ – Liddo Dec 5 '18 at 16:53
  • 1
    $\begingroup$ It is of interest to know that Ramanujan proved this congruence using the same method. He provided another more complicated proof by giving a closed form for $\sum_{n=0}^{\infty} p(7n+5)q^n$. See this post for more details. $\endgroup$ – Paramanand Singh Dec 6 '18 at 17:13

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 $$[q^{7n+7}]\frac{q^{2}((q;q)^{3}_{\infty})^{2}}{(q^{7};q^{7})_{\infty}} = [q^{7n+7}](q^{2}((q;q)^{3}_{\infty})^{2})\sum_{i=0}^\infty P(i)q^{7i} = \sum_{i=0}^{n+1} P(i)[q^{7(n-i+1)}](q^{2}((q;q)^{3}_{\infty})^{2})$$


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