[proofplan] The proof uses Euler's product for $\zeta$ in the half-plane $\operatorname{Re}(s)>1$ and the positivity identity $3+4\cos u+\cos 2u=2(1+\cos u)^2\geq 0$. For $\sigma>1$ we convert this positivity into the lower bound $|\zeta(\sigma)^3\zeta(\sigma+it)^4\zeta(\sigma+2it)|\geq 1$. If $\zeta(1+it)=0$ with $t\neq 0$, the simple pole of $\zeta$ at $1$ contributes order $-3$, while the assumed zero contributes order at least $4$; the remaining factor is holomorphic at $1+2it$. Hence the same product tends to $0$ as $\sigma\downarrow 1$, contradicting the lower bound. [/proofplan] [step:Convert trigonometric positivity into a zeta-product lower bound] Fix $t\in\mathbb{R}\setminus\{0\}$. For $\sigma>1$, define \begin{align*} P_t:(1,\infty)&\to (0,\infty)\\ \sigma&\mapsto \left|\zeta(\sigma)^3\zeta(\sigma+it)^4\zeta(\sigma+2it)\right|. \end{align*} We use the Euler product logarithm for the Riemann zeta function in the half-plane $\operatorname{Re}(s)>1$: \begin{align*} \log \zeta(s)=\sum_{p}\sum_{k=1}^{\infty}\frac{p^{-ks}}{k}, \end{align*} where the outer sum is over all primes $p$, and the series is absolutely convergent for $\operatorname{Re}(s)>1$ (citing a result not yet in the wiki: Euler product and logarithm for the Riemann zeta function). Taking real parts gives, for each real $a$ and each $\sigma>1$, \begin{align*} \log |\zeta(\sigma+ia)| &=\operatorname{Re}\log \zeta(\sigma+ia)\\ &=\sum_{p}\sum_{k=1}^{\infty}\frac{\cos(ka\log p)}{k p^{k\sigma}}. \end{align*} Therefore \begin{align*} \log P_t(\sigma) &=3\log|\zeta(\sigma)|+4\log|\zeta(\sigma+it)|+\log|\zeta(\sigma+2it)|\\ &=\sum_{p}\sum_{k=1}^{\infty} \frac{3+4\cos(kt\log p)+\cos(2kt\log p)}{k p^{k\sigma}}. \end{align*} For every real number $u$, \begin{align*} 3+4\cos u+\cos 2u &=3+4\cos u+2\cos^2u-1\\ &=2(1+\cos u)^2\geq 0. \end{align*} Applying this with $u=kt\log p$ shows that every summand in the double series for $\log P_t(\sigma)$ is non-negative. Hence $\log P_t(\sigma)\geq 0$, and therefore \begin{align*} P_t(\sigma)\geq 1 \end{align*} for every $\sigma>1$. [guided] Fix $t\in\mathbb{R}\setminus\{0\}$. The goal is to build a quantity involving $\zeta(\sigma)$, $\zeta(\sigma+it)$, and $\zeta(\sigma+2it)$ whose logarithm is visibly non-negative. Define \begin{align*} P_t:(1,\infty)&\to (0,\infty)\\ \sigma&\mapsto \left|\zeta(\sigma)^3\zeta(\sigma+it)^4\zeta(\sigma+2it)\right|. \end{align*} For $\operatorname{Re}(s)>1$, Euler's product gives an absolutely convergent logarithmic expansion \begin{align*} \log \zeta(s)=\sum_{p}\sum_{k=1}^{\infty}\frac{p^{-ks}}{k}, \end{align*} where $p$ runs over the primes (citing a result not yet in the wiki: Euler product and logarithm for the Riemann zeta function). Absolute convergence justifies taking real parts term-by-term. If $a\in\mathbb{R}$ and $\sigma>1$, then \begin{align*} p^{-k(\sigma+ia)} =p^{-k\sigma}e^{-ika\log p}, \end{align*} so \begin{align*} \operatorname{Re}\left(p^{-k(\sigma+ia)}\right) =p^{-k\sigma}\cos(ka\log p). \end{align*} Thus \begin{align*} \log |\zeta(\sigma+ia)| &=\operatorname{Re}\log \zeta(\sigma+ia)\\ &=\sum_{p}\sum_{k=1}^{\infty}\frac{\cos(ka\log p)}{k p^{k\sigma}}. \end{align*} Now substitute $a=0$, $a=t$, and $a=2t$, with coefficients $3$, $4$, and $1$. This gives \begin{align*} \log P_t(\sigma) &=3\log|\zeta(\sigma)|+4\log|\zeta(\sigma+it)|+\log|\zeta(\sigma+2it)|\\ &=\sum_{p}\sum_{k=1}^{\infty} \frac{3+4\cos(kt\log p)+\cos(2kt\log p)}{k p^{k\sigma}}. \end{align*} The specific coefficients $3,4,1$ were chosen because the trigonometric polynomial is non-negative. Indeed, for every $u\in\mathbb{R}$, \begin{align*} 3+4\cos u+\cos 2u &=3+4\cos u+2\cos^2u-1\\ &=2(1+\cos u)^2\geq 0. \end{align*} With $u=kt\log p$, every term of the double series for $\log P_t(\sigma)$ is non-negative. Since the series converges absolutely for $\sigma>1$, its sum is non-negative: \begin{align*} \log P_t(\sigma)\geq 0. \end{align*} Exponentiating preserves the inequality, so \begin{align*} P_t(\sigma)\geq 1 \end{align*} for every $\sigma>1$. [/guided] [/step] [step:Estimate the same product near the line from a hypothetical zero] Assume, for contradiction, that $\zeta(1+it)=0$. Let $m\in\mathbb{N}$ denote the order of this zero. Since $\zeta$ is holomorphic at $1+it$ for $t\neq 0$, there exist $\varepsilon_1>0$ and $C_1>0$ such that \begin{align*} |\zeta(\sigma+it)|\leq C_1(\sigma-1)^m \end{align*} whenever $1<\sigma<1+\varepsilon_1$. The Riemann zeta function has a simple pole at $s=1$ (citing a result not yet in the wiki: meromorphic continuation of $\zeta$ and its simple pole at $1$). Hence there exist $\varepsilon_2>0$ and $C_2>0$ such that \begin{align*} |\zeta(\sigma)|\leq C_2(\sigma-1)^{-1} \end{align*} whenever $1<\sigma<1+\varepsilon_2$. Since $t\neq 0$, the point $1+2it$ is not equal to $1$. The meromorphic continuation of $\zeta$ is holomorphic at $1+2it$, so there exist $\varepsilon_3>0$ and $C_3>0$ such that \begin{align*} |\zeta(\sigma+2it)|\leq C_3 \end{align*} whenever $1<\sigma<1+\varepsilon_3$. Let $\varepsilon=\min\{\varepsilon_1,\varepsilon_2,\varepsilon_3\}$ and define $C=C_1^4C_2^3C_3>0$. For $1<\sigma<1+\varepsilon$, \begin{align*} P_t(\sigma) &=\left|\zeta(\sigma)^3\zeta(\sigma+it)^4\zeta(\sigma+2it)\right|\\ &\leq C_2^3(\sigma-1)^{-3}\, C_1^4(\sigma-1)^{4m}\, C_3\\ &=C(\sigma-1)^{4m-3}. \end{align*} Since $m\geq 1$, we have $4m-3\geq 1$, and therefore \begin{align*} \lim_{\sigma\downarrow 1}P_t(\sigma)=0. \end{align*} [guided] Assume, for contradiction, that $\zeta(1+it)=0$. Because $t\neq 0$, the point $1+it$ is not the pole $s=1$, so $\zeta$ is holomorphic at $1+it$. Let $m\in\mathbb{N}$ be the order of the zero at $1+it$. By the definition of order of vanishing, there is a [holomorphic function](/page/Holomorphic%20Function) $h$ in a neighbourhood of $1+it$ with $h(1+it)\neq 0$ and \begin{align*} \zeta(s)=(s-(1+it))^m h(s) \end{align*} near $1+it$. Restricting to $s=\sigma+it$ gives \begin{align*} |\zeta(\sigma+it)|=(\sigma-1)^m |h(\sigma+it)|. \end{align*} Since $h$ is continuous near $1+it$, it is bounded on a sufficiently small closed neighbourhood. Hence there exist $\varepsilon_1>0$ and $C_1>0$ such that \begin{align*} |\zeta(\sigma+it)|\leq C_1(\sigma-1)^m \end{align*} whenever $1<\sigma<1+\varepsilon_1$. Next we estimate the factor at $\sigma$. The Riemann zeta function has a simple pole at $s=1$ (citing a result not yet in the wiki: meromorphic continuation of $\zeta$ and its simple pole at $1$). Therefore there is a holomorphic function $g$ near $1$ with $g(1)\neq 0$ and \begin{align*} \zeta(s)=\frac{g(s)}{s-1} \end{align*} near $s=1$. Restricting to $s=\sigma$ and using boundedness of $g$ on a small neighbourhood gives constants $\varepsilon_2>0$ and $C_2>0$ such that \begin{align*} |\zeta(\sigma)|\leq C_2(\sigma-1)^{-1} \end{align*} whenever $1<\sigma<1+\varepsilon_2$. Finally, because $t\neq 0$, the point $1+2it$ is not $1$. The only pole of $\zeta$ on this part of the line is at $1$, so $\zeta$ is holomorphic at $1+2it$ (citing a result not yet in the wiki: meromorphic continuation of $\zeta$ with unique simple pole at $1$). A holomorphic function is locally bounded, so there exist $\varepsilon_3>0$ and $C_3>0$ such that \begin{align*} |\zeta(\sigma+2it)|\leq C_3 \end{align*} whenever $1<\sigma<1+\varepsilon_3$. Let $\varepsilon=\min\{\varepsilon_1,\varepsilon_2,\varepsilon_3\}$ and define \begin{align*} C=C_1^4C_2^3C_3. \end{align*} For $1<\sigma<1+\varepsilon$, multiplying the three estimates gives \begin{align*} P_t(\sigma) &=\left|\zeta(\sigma)^3\zeta(\sigma+it)^4\zeta(\sigma+2it)\right|\\ &\leq C_2^3(\sigma-1)^{-3}\, C_1^4(\sigma-1)^{4m}\, C_3\\ &=C(\sigma-1)^{4m-3}. \end{align*} The exponent is positive because $m\geq 1$, so $4m-3\geq 1$. Hence \begin{align*} \lim_{\sigma\downarrow 1}P_t(\sigma)=0. \end{align*} This is the order-counting heart of the proof: the assumed zero contributes at least four powers of $\sigma-1$, while the pole at $1$ can remove only three. [/guided] [/step] [step:Derive the contradiction and handle the point $s=1$] The first step proves $P_t(\sigma)\geq 1$ for every $\sigma>1$. The second step, under the assumption $\zeta(1+it)=0$, proves \begin{align*} \lim_{\sigma\downarrow 1}P_t(\sigma)=0. \end{align*} These two conclusions are incompatible. Therefore $\zeta(1+it)\neq 0$ for every $t\in\mathbb{R}\setminus\{0\}$. At $t=0$, the point on the line is $s=1$. The Riemann zeta function has a simple pole at $s=1$, so $s=1$ is not a zero. Hence $\zeta$ has no zeros on the line $\operatorname{Re}(s)=1$. [/step]