Ytukyg

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

$displaystylesum_{ngeq0};P(n)q^{n}=frac{1}{(q;q)_{infty}}$

we multiply by $q^{2}$ then

begin{eqnarray}
displaystylesum_{ngeq0};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 $(q;q)^{3}_{infty}=displaystylesum_{ngeq0} (-1)^{n}(2n+1)q^{frac{n(n+1)}{2}}$, 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 $$(2n+1)^{2}+(2m+1)^{2}=8left(frac{n(n+1)}{2}+frac{m(m+1)}{2}+2right)-14$$

Then $frac{n(n+1)}{2}+frac{m(m+1)}{2}+2equiv 0(mod;7)$ 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 $2n+1equiv0(mod ;7)$ and $2m+1equiv0(mod ;7)$.

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 $$frac{1}{(q;q)^{7}_{infty}}equivfrac{1}{(q^{7};q^{7})_{infty}}(mod; 7 )$$

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