[Math] Why does the Riemann zeta function have non-trivial zeros

analytic-number-theorycv.complex-variablesnt.number-theorypolymath5zeta-functions

This is a very basic question of course, and exposes my serious ignorance of analytic number theory, but what I am looking for is a good intuitive explanation rather than a formal proof (though a sufficiently short formal proof could count as an intuitive explanation). So, for instance, a proof that estimated a contour integral and thereby showed that the number of zeros inside the contour was greater than zero would not count as a reason. A brief glance at Wikipedia suggests that the Hadamard product formula could give a proof: if there are no non-trivial zeros then you get a suspiciously nice formula for ζ(s) itself. But that would feel to me like formal magic. A better bet would probably be Riemann's explicit formula, but that seems to require one to know something about the distribution of primes. Perhaps a combination of the explicit formula and the functional equation would do the trick, but that again leaves me feeling as though something magic has happened. Perhaps magic is needed.

A very closely related question is this. Does the existence of non-trivial zeros on the critical strip imply anything about the distribution of prime numbers? I know that it implies that the partial sums of the Möbius and Liouville functions cannot grow too slowly, and it's really this that I want to understand.

Best Answer

If the Riemann zeta function had only trivial zeroes, then after multiplying by the gamma factor, it would become a zero-free entire function of finite order. Every zero-free entire function of finite order is of the form $\exp(poly(z))$. This cannot happen for the zeta function, as can be seen by considering behavior as $s\to\infty$ along the reals, for example.

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