\(\newcommand{\AA}{\mathcal{A}} \newcommand{\CH}{\mathcal{H}} \newcommand{\CL}{\mathcal{L}} \newcommand{\CO}{\mathcal{O}} \newcommand{\FF}{\mathbb{F}} \newcommand{\NN}{\mathbb{N}} \newcommand{\CC}{\mathbb{C}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\ZZ}{\mathbb{Z}}\) \(\DeclareMathOperator{\an}{an} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\colim}{colim} \DeclareMathOperator{\Hom}{Hom} \DeclareMathOperator{\Id}{Id} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\Isom}{Isom} \DeclareMathOperator{\Gal}{Gal} \DeclareMathOperator{\GL}{GL} \DeclareMathOperator{\Ob}{Ob} \DeclareMathOperator{\Res}{Res} \DeclareMathOperator{\sh}{sh} \DeclareMathOperator{\SL}{SL} \DeclareMathOperator{\Spec}{Spec} \DeclareMathOperator{\Sym}{Sym}\)
Lemma 3.1.   Ramanujan's conjecture is true if the roots of \(H_p\) are of absolute value \(p^{11/2}\).

Proof. We see \(H_p\) has roots

\[\alpha_{\pm} = \frac{\tau(p) \pm \sqrt{\tau(p)^2 - 4 p^{11}}}{2}\]

Hence, if

\[2 p^{11/2} = \left| \tau(p) \pm \sqrt{\tau(p)^2 - 4 p^{11}} \right|,\]

choosing our sign \(\pm\) appropriately shows \(\vert \tau(p) \vert \leq 2 p^{11/2}.\) \(\blacksquare\)