 Integrality identity for the modular discriminant

Put $S_j={\sum }_{n\ge 1}{\sigma }_j(n)q^n$. Expanding $E_4=1+240S_3$ and $E_6=1-504S_5$ gives $\Delta =(5S_3+7S_5)/12+100S_3^2+8000S_3^3-147S_5^2$. For every integer $d$, $5d^3+7d^5$ is divisible by twelve, by separate congruences modulo three and four. Consequently every coefficient of the modular discriminant is integral, without using a product expansion.