Krdytkyu

4 $begingroup$ 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...