Write
$$
a_1(E_k)=\frac{2}{\zeta(1-k)}=\frac rs,
\qquad \gcd(r,s)=1,
$$
and suppose $p\mid s$. In the basis from part b, comparison of the constant term and the first $N$ nonconstant coefficients gives
$$
sE_k=sf_0+r\sum_{i=1}^N\sigma_{k-1}(i)f_i.
$$
Indeed, $f_0$ has constant term one and no terms $q,ldots,q^N$, while $f_i$ has the sole term $q^i$ in that range.

Set
$$
f=\sum_{i=1}^N\sigma_{k-1}(i)f_i.
$$
Every $f_i$ with $i\geq1$ vanishes at infinity and is therefore a <cusp form>; moreover $f$ has integral coefficients. Comparing the coefficient of $q^n$ in the displayed identity gives
$$
r\sigma_{k-1}(n)=s,a_n(f_0)+r,a_n(f).
$$
Reduction modulo $p$ kills the first term on the right. Since $p\nmid r$, cancellation of $r$ yields
$$
a_n(f)\equiv\sigma_{k-1}(n)\pmod p
$$
for every $n\geq1$, proving the <Eisenstein congruence from a denominator prime>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 137 / 1 / c / Solution

If the first nonconstant coefficient of $E_k$ has reduced denominator divisible by a prime $p$, the integral echelon basis produces an integral cusp form whose Fourier coefficients are congruent modulo $p$ to $\sigma_{k-1}(n)$.


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 Incoming links: Eisenstein congruence from a denominator prime