Rather than the problem of why the zeta function has non-trivial zeros, let me address
Gowers's question of why the error term in the prime number theorem needs to be large. The short answer that I propose is: *because the integers are so well distributed*. To make this precise, I shall prove a general result on semigroups, showing that
either the "integers" in the semigroup or the "primes" must be poorly distributed -- one may think of this as an "uncertainty principle" that both primes and integers cannot be simultaneously smoothly behaved. This has the flavor of Beurling generalized primes, but I don't recall this result in the literature; maybe it exists already (indeed it does, see the edit below). Also note that the proof will not make any use
of the functional equation for $\zeta(s)$ as this does not exist in a general semigroup.

EDIT: Indeed doing a literature search a few hours after posting this,
I found a paper of Hilberdink http://www.sciencedirect.com/science/article/pii/S0022314X04002069 which proves the
Theorem below, and with a similar method of proof.

Suppose that $1<p_1< p_2 <\ldots$ is a sequence of real numbers (the "primes"), and
$1=a_1 \le a_2 \le \ldots$ is the semigroup generated by them (the "integers"). Initially I made the assumption that the "integers" are distinct, and satisfy a mild spacing
condition $a_{n+1}-a_n \gg n^{-1}$ (so that in particular there is unique factorization), but this is not necessary. Let $N(x)$ denote the number of
"integers" below $x$, and
$$
P(x) = \sum_{p_n^{k} \le x} \log p_n,
$$
which is the analog of the usual $\psi(x)$ (counting prime powers with weight $\log p$). Assume that for some $\delta>0$
$$
N(x) = Ax +O(x^{\frac 12-\delta}),
$$
for some non-zero constant $A$, and that
$$
P(x) = x + O(x^{\frac 12-\delta}).
$$

**Theorem**. *Either the asymptotic formula for $N(x)$ or the asymptotic
formula for $P(x)$ must fail.*

Put
$$
\zeta_A(s) = \sum_{n=1}^{\infty} a_n^{-s} = \prod_{n} \Big(1-\frac{1}{p_n^s}\Big)^{-1}.
$$

By our assumptions on $N$ and $P$, the sum and product above converge absolutely in Re$(s)>1$. By the assumption on $N(x)$, $\zeta_A(s)$ extends to an analytic function in Re$(s)>1/2-\delta$ except for a simple pole at $s=1$ with residue $A$. By the assumption on $P$, we see that the logarithmic derivative
$$
-\frac{\zeta_A^{\prime}}{\zeta_A}(s) = \sum_{n, k} \frac{\log p_n}{p_n^{ks}}
$$
extends analytically to Re$(s)>1/2-\delta$ except for a simple pole at $1$. Thus
$\zeta_A(s)$ has no zeros in Re$(s)>1/2-\delta$.

**New edit: Sketch of a second proof.** Adapting the argument that the Riemann hypothesis implies the Lindelof hypothesis (see below), we obtain that $|\zeta_A(s)| \ll (1+|s|)^{\epsilon}$ provided $s$ is not close to the pole at $1$, and that Re$(s)>1/2-\delta/2$. From this and a standard contour shift argument we find that for large $N$ and any $t$,
$$
\sum_{a_n\le N} a_n^{it} = \frac{N^{1+it}}{1+it} +O(N^{1/2-\delta+\epsilon} (1+|t|)^{\epsilon}).
$$
What is used here is that we have Lindelof even a little to the left of the half line.

But the above identity can be seen to contradict the Plancherel formula. More precisely, let $T$ be a large power of $N$, and let $\Phi$ be a non-negative function supported in $[-1,1]$ with non-negative Fourier transform. Then we see that (discarding all but the diagonal terms)
$$
\int_{-\infty}^{\infty} \Big| \sum_{a_n\le N} a_n^{it}\Big|^2 \Phi(t/T) dt
\ge T{\hat \Phi}(0) \sum_{a_n\le N} 1 \sim TN {\hat \Phi}(0).
$$
On the other hand, if we use our identity then the above is seen to be
$$
\ll N^2 + T^{1+\epsilon} N^{1-2\delta}.
$$
This is a contradiction.

**Original Proof:** Below let's assume always that we are in the region Re$(s)>1/2-\delta$, and that
the imaginary part is large so that we are not near the pole at $1$. From
the analytic continuation of $\zeta_A$ (using that $N(x)$ is very regular), it follows
that there is an a priori polynomial bound $|\zeta_A(s)|\ll |s|^{B}$ in the region Re$(s)>1/2-\delta/2$. Thus there is a bound for the real part of $\log \zeta_{A}(s)$, and by the Borel-Caratheodory lemma (standard complex analysis) one can bootstrap this to a bound for $|\log \zeta_A(s)|$. Then applying the Hadamard three circle theorem to $\log \zeta_A(s)$ one obtains a much better bound: $|\zeta_A(s)| \ll |s|^{\epsilon}$. This is the usual proof that Riemann implies Lindelof. (At this stage, if we knew a "functional equation" we'd be done, as the usual $\zeta(s)$ is large when Re$(s)<1/2$. This point appeared in Matt Young's answers earlier.)

Knowing the Lindelof hypothesis for $\zeta_A(s)$ in Re$(s)> 1/2-\delta/2$, we can
show the following approximate formula: for $\sigma > 1/2- \delta/4$
$$
\zeta_A(\sigma +it) = \sum_{a_n \le N} a_n^{-\sigma-it} + O(|t|^{\epsilon} N^{-\delta/4}).
$$
The proof is standard; see the penultimate chapter of Titchmarsh for the real $\zeta(s)$ where this holds when $\sigma$ is strictly bigger than $1/2$, and our stronger result is true because we have Lindelof in a wider region.

Now we are ready to get our contradiction. Consider for large $T$
$$
\int_T^{2T} |\zeta_A(1/2+it)|^2 dt.
$$
To do this carefully it may be helpful to put in a smooth weight $\Phi(t/T)$ above
(but this is not a paper!). Using the approximate formula derived above, and our mild spacing condition $a_{n+1}-a_n \gg n^{-1}$,
we may see that for any $T^{\epsilon} \le N \le T^{1/10}$ we have
$$
\int_T^{2T} |\zeta_A(1/2+it)|^2 dt \sim T \sum_{a_n \le N} a_n^{-1} \sim AT \log N.
$$
But that's absurd! This completes our sketch proof.

in an elementary waythat the summatory function of $\mu(n)$ is unbounded. It seems that complex analysis is quite difficult to dispense with in these kinds of problems. Unfortunately, analytic arguments do not seem to give anyarithmetical insightinto the reasons for comparative largeness of the error terms. $\endgroup$7more comments