Trivial Zeroes of Riemann Zeta Function are Even Negative Integers

Theorem
Suppose $\rho$ is a zero of the Riemann zeta function not contained in the critical strip:
 * $0 \le \Re \left({s}\right) \le 1$

Then:
 * $s \in \left\{{-2, -4, -6, \ldots}\right\}$

These are called the trivial zeros of $\zeta$.

Proof
First we note that by Zeroes of the Gamma Function, $\Gamma$ has no zeros on $\C$.

Therefore, the completed Riemann zeta function:


 * $\displaystyle \xi \left({s}\right) = \frac 1 2 s \left({s - 1}\right) \pi^{-s/2} \Gamma \left({\frac s 2}\right) \zeta \left({s}\right)$

has the same zeros as $\zeta$.

Additionally by Functional Equation for Riemann Zeta Function, we have $\xi \left({s}\right) = \xi \left({1-s}\right)$ for all $s \in \C$.

Therefore if $\zeta \left({s}\right) \ne 0$ for all $s$ with $\Re \left({s}\right) > 1$ then also $\zeta \left({s}\right) \ne 0$ for all $s$ with $\Re \left({s}\right) < 0$.

Let us consider $\Re \left({s}\right) > 1$. We have:


 * $\displaystyle \zeta \left({s}\right) = \prod_p \frac 1 {1 - p^{-s}}$

where here and in the following $p$ ranges over the primes.

Therefore, we have:
 * $\displaystyle \zeta \left({s}\right) \prod_p \left({1 - p^{-s}}\right) = 1$

All of the factors of this infinite product can be found in the product:


 * $\displaystyle \prod_{n=2}^\infty \left({1 - n^{-s}}\right)$

which converges absolutely since the zeta sum $\displaystyle \sum_{k=1}^\infty k^{-s}$ converges absolutely.

Hence:


 * $\displaystyle \prod_p \left({1 - p^{-s}}\right)$

converges absolutely, and so by the fact that:


 * $\displaystyle \zeta \left({s}\right) \prod_p \left({1-p^{-s}}\right) = 1$

we know $\zeta \left({s}\right)$ can't possibly be zero for any point in the region in question.