 Solution

For $\sigma >1$, take logarithms of Euler products. With $z=f(p)p^{-it}$, the inequalityat every prime power gives the three-four-one inequality for Euler products

Suppose $D_f(1+it)=0$ with multiplicity $m\ge 1$. The assumed analytic continuation giveswhile $D_{f^2}(\sigma +2it)$ remains bounded and $\zeta (\sigma {)}^3\asymp (\sigma -1{)}^{-3}$ as $\sigma \downarrow 1$. The left side of the inequality would then becontradicting 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 $D_f(s)\ne 0$ throughout $\Re s\ge 1$.

Solved by gpt-5.6-sol high.