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

Solution

Write

a_{1} (E_{k}) = \frac{2}{ζ ( 1 - k )} = \frac{r}{s}, g cd (r, s) = 1,

and suppose

p ∣ s

. In the basis from part b, comparison of the constant term and the first

N

nonconstant coefficients gives

s E_{k} = s f_{0} + r \sum_{i = 1}^{N} σ_{k - 1} (i) f_{i} .

Indeed,

f_{0}

has constant term one and no terms

q, l d o t s, q^{N}

, while

f_{i}

has the sole term

q^{i}

in that range.

f = \sum_{i = 1}^{N} σ_{k - 1} (i) f_{i} .

Every

f_{i}

with

i \geq 1

vanishes at infinity and is therefore a cusp form; moreover

f

q^{n}

in the displayed identity gives

r σ_{k - 1} (n) = s, a_{n} (f_{0}) + r, a_{n} (f) .

Reduction modulo

p

kills the first term on the right. Since

p ∤ r

, cancellation of

r

yields

a_{n} (f) \equiv σ_{k - 1} (n) (mod p)

for every

n \geq 1

There are no discussions about this article yet.

There are currently no matching articles.