= Solution
The preceding <dimension of level-one modular forms> gives $\dim M_{16}=2$ and $\dim S_{16}=\dim M_4=1$. Thus $E_4\Delta$, which begins with $q$, spans $S_{16}$. A normalized weight-sixteen <Eisenstein series> and $E_4^4$ both have constant term one, so their difference is a multiple of $E_4\Delta$.
There is a normalization error in the printed auxiliary expansion: for the usual normalized $E_{16}$, the coefficient multiplier is $16320/3617$, rather than $255/(8\cdot3617)$. Indeed, the <Fourier expansion of a normalized Eisenstein series> gives $-32/B_{16}$, and the <Bernoulli number> $B_{16}=-3617/510$ gives $16320/3617$. The printed multiplier is smaller by a factor of $512$. With the correct normalization, the first <Fourier coefficients> yield
$$
E_4^4=1+960q+O(q^2),\qquad E_{16}=E_4^4-\frac{3456000}{3617}E_4\Delta.
$$
Comparing all positive-index coefficients therefore gives the exact identity
$$
16320\,\sigma_{15}(n)=3617\,[q^n]E_4^4-3456000\,c_n.
$$
The <Fourier coefficients> of $E_4^4$ are integers, as are those of $E_4\Delta$: $E_4$ has its integral divisor-sum expansion and the <modular discriminant> has the product $q\prod_{r\geq1}(1-q^r)^{24}$. Also $3456000+16320=960\cdot3617$, and $\gcd(16320,3617)=1$. Reducing the identity modulo $3617$ and cancelling the invertible multiplier proves the <weight-sixteen Eisenstein congruence>
$$
\boxed{c_n\equiv\sigma_{15}(n)\pmod{3617}\qquad(n\geq1).}
$$
The desired congruence is valid; it is the auxiliary normalization in the PDF that requires correction.
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