 Ramanujan tau congruence modulo five

ID: ramanujan-tau-congruence-modulo-five

1728Δ = E_{4}^{3} - E_{6}^{2}

, while the Serre derivative gives

2Θ E_{6} = E_{2} E_{6} - E_{4}^{2}

. Their integral Fourier expansions have

E_{4} \equiv 1

and

E_{2} \equiv E_{6} (mod 5)

, because

d^{5} \equiv d (mod 5)

. Reducing the two identities gives

Δ \equiv Θ E_{6} (mod 5)

q^{n}

Θ E_{6}

- 504 n σ_{5} (n) \equiv n σ_{5} (n)

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