= Solution
For $\sigma>1$, take logarithms of Euler products. With $z=f(p)p^{-it}$, the inequality
$$
3+4\Re(z^k)+\Re(z^{2k})\geq0
\qquad(|z|\leq1)
$$
at every prime power gives the <three-four-one inequality for Euler products>
$$
\zeta(\sigma)^3|D_f(\sigma+it)|^4|D_{f^2}(\sigma+2it)|\geq1.
$$
Suppose $D_f(1+it)=0$ with multiplicity $m\geq1$. The assumed analytic continuation gives
$$
|D_f(\sigma+it)|\ll(\sigma-1)^m,
$$
while $D_{f^2}(\sigma+2it)$ remains bounded and $\zeta(\sigma)^3\asymp(\sigma-1)^{-3}$ as $\sigma\downarrow1$. The left side of the inequality would then be
$$
O((\sigma-1)^{4m-3})\longrightarrow0,
$$
contradicting its lower bound one. Hence $D_f$ has no zero on $\Re s=1$. Absolute convergence of its Euler product already excludes zeros for $\Re s>1$, so $D_f(s)\ne0$ throughout $\Re s\geq1$.
Solved by gpt-5.6-sol high.
Back to article page
