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I have trouble trying to prove that the cusp form $\Delta$ has always integer coefficients. Using \begin{equation} \Delta = (E_4^3-E_6^2)/1728 \end{equation} And knowing that \begin{equation} E_4 = 1-\frac{8}{B_4}q-\frac{8}{B_4}\sum_{n\geq 2}\sigma_3(n)q^n \end{equation}

\begin{equation} E_6 = 1-\frac{12}{B_6}q-\frac{12}{B_6}\sum_{n\geq 2}\sigma_5(n)q^n \end{equation}

I calculated $E_4^3-E_6^2$ by hand and the first coefficient of $q$ is indeed $1728$. The coefficiente of $q^2$ is $-41472$ which is divisible by $1728$. I also calculated the coefficient of $q^3$ ($435456$) but for $q^k$, $k\geq 4$ it is not obvious if its coefficient is divisible by $1728$ without evaluating the expression explicitly. What can I do?

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  • $\begingroup$ You could try to the Jacoi product factorization, which clearly has integral coefficients when expanded. $\endgroup$ – KCd Jun 9 '17 at 3:19
  • $\begingroup$ You can see this answer (math.stackexchange.com/a/2327018/72031) where I directly prove that $$E_{4}^{3}-E_{6}^{2}=1728q\prod_{n=1}^{\infty}(1-q^{n})^{24}$$ so that $\Delta $ has integer coefficients. $\endgroup$ – Paramanand Singh Sep 9 '17 at 7:11
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It is well known that $$\Delta=q\prod_{n=1}^\infty(1-q^n)^{24}.$$

On the other hand $$E_4=1+240\sum_{n=1}^\infty \sigma_3(n) q^n=1+480f(q)$$ and $$E_6=1-504\sum_{n=1}^\infty \sigma_5(n) q^n=1-504g(q)$$ where $f$ and $g$ have integer coefficients. Then $$E_4^3-E_6^2=720f(q)+172800f(q)^2+13824000f(q)^3 +1008g(q)-254016g(q)^2 \equiv 720f(q)+1008g(q)=144(5f(q)+7g(q))\pmod{1728}.$$ We need to prove that $5f(q)+7g(q)\equiv0\pmod{12}$.

The $q^n$ coefficient of $5f(q)+7g(q)$ is $$5\sigma_3(n)+7\sigma_5(n)=\sum_{d\mid n} (5d^3+7d^5).$$ All that remains to to is to prove that $5d^3+7d^5$ is always a multiple of $12$.

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