Past exam of the analytic number theory and elliptic curves course of the University of Oxford / 2007 / A / 3

The double-angle formula givesFor $\sigma >1$, the logarithm of the absolutely convergent Euler product givesExponentiation provesSuppose $\zeta (1+it)=0$ for $t\ne 0$, of order $r\ge 1$. As $\sigma \downarrow 1$, its middle factor is $O((\sigma -1{)}^{4r})$, the first is $O((\sigma -1{)}^{-3})$ because the pole at one is simple, and the final factor is bounded because $1+2it\ne 1$. The product is $O((\sigma -1{)}^{4r-3})\to 0$, contradicting the lower bound one. ThereforeThis is the three-four-one zero-free-region argument specialized to the line with real part one.

Complete multiplicativity and $f(p)=\pm 1$ give, for $\mathrm{Re}s>1$,The product is nonzero there; its absolutely convergent logarithm has terms $f(p{)}^m{p}^{-ms}/m$. ThusEach numerator is one of $3\pm 4\cos \theta +\cos 2\theta$, so it is nonnegative. Exponentiation gives the required lower bound one.

If $F(1+it)=0$ with $t\ne 0$, its assumed holomorphy makes $|F(\sigma +it){|}^4=O((\sigma -1{)}^4)$, or smaller for a higher-order zero. The zeta factor at real $\sigma$ grows at most as $(\sigma -1{)}^{-3}$, while the zeta factor near $1+2it$ stays bounded. The product tends to zero, a contradiction. Hence