 Solution

The normalized Eisenstein series satisfy $E_4=1+240{\sum }_{m\ge 1}{\sigma }_3(m)q^m$ and $E_6=1-504{\sum }_{m\ge 1}{\sigma }_5(m)q^m$. Their weights are four and six. The valence formula for the modular group makes the normalized weight-twelve cusp form $\Delta$ have one simple cusp zero and no zero in $\mathbb{H}$. Comparing its first coefficient gives $\Delta =(E_4^3-E_6^2)/1728$.

Odd weights vanish because $-I$ acts by $(-1{)}^k$. Weight two vanishes by the valence formula for the modular group: the forced zero at $i$ alone contributes $1/2>2/12$. For every remaining nonnegative even weight, except two, choose $a,b\ge 0$ with $4a+6b=k$. Subtract the constant coefficient of $f$ times $E_4^a{E}_6^b$. The remainder is a cusp form; division by $\Delta$ produces a holomorphic modular form of weight $k-12$. The cusp and interior orders justify this division. Induction on the weight provesThe integral triangular basis of level-one modular forms is $F_j={\Delta }^j{E}_4^{a_j}{E}_6^{b_j}$, with $4a_j+6b_j=k-12j\ge 0$ and $k-12j\ne 2$. The induction just given shows these span, and their distinct leading terms $q^j$ show linear independence. Every coefficient is integral by the hypothesis on $\Delta$ and the displayed Eisenstein expansions. In $f={\sum }_j{c}_j{F}_j$, compare coefficients successively: $c_0=a_0$, then $c_1$ is $a_1$ minus an integer combination of already integral coefficients, and similarly for every $j$. The largest permitted $j$ is at most $\lfloor (k+1)/12\rfloor$, so the stipulated initial coefficients make every $c_j$ integral. Therefore all Fourier coefficients of $f$ are integers.