Riemann hypothesis and diophantine equation

I estimate that with the current state of knowledge, writing the Riemann hypothesis in true Diophantine terms would require equations with over 100 variables.

Statement of the Riemann Hypothesis in integers only

The Riemann hypothesis is equivalent to a lack of solutions for \begin{align} (kq)^k s +a&= (kq+p)^k t\ \ \ \ \ \ \ \ \ \ (i)\\ k^k p^s s^t +b&= (ct+k)^{k} q^s t^t\ \ \ \ \ \ (ii)\\ s &= \prod_{i=1}^n (2mi+1)^m\ \ \ (iii)\\ t &= \prod_{i=1}^n (2mi-1)^m\ \ \ (iv)\\ c &= \sum_{d \mid n} d \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (v)\\ \end{align} where $a,b,c,k,m,n,p,q,s,t$ are positive integers. This gets the question entirely into the realm of integers, with no computation of rationals or reals required, and is probably a reasonable stopping point for explicitness.

Estimates for statement of Riemann hypothesis in polynomials only

These could be converted into Diophantine equations, i.e. equations of polynomials, by the techniques in Davis's exposition of Hilbert's tenth problem.

• For equations $(i)$ and $(ii)$, each exponential abbreviates a set of polynomial equations as in lemma 3.5, with 19 new variables.
• For equations $(iii)$ and $(iv)$, each product of linear terms abbreviates a set of polynomial equations as in lemma 4.7, with an estimated 215 new variables.
• For equation $(v)$, the sum requires the $m=12$, $n=2$ case of lemma 5.2, including 13 products of linear terms, and an estimated total of 2947 variables.

In all I estimate 3567 variables needed to translate the above equations into pure Diophantine form by Davis's procedures. These can obviously be reduced, since Davis was not optimizing for number of variables but for clarity of exposition. However, even reducing the requirements a hundredfold would still probably yield something unreadable.

Proof of equivalence for above statement of Riemann Hypothesis

To see where equations $(i)-(v)$ come from, note that $$\frac{(2mi+1)^m}{(2mi-1)^m} \text{ and } \frac{(kq+p)^k}{(kq)^k}$$ are upper and lower approximations to $\exp(1/i)$ and $\exp(p/q)$. In particular $$\lim_{m\rightarrow\infty}\frac{s}{t}=\exp\left(1+\frac{1}{2}+\cdots\dfrac{1}{n}\right)=\exp(H_n)$$ with $s/t > \exp(H_n)$.

Now suppose we had a solution to the equations. Then equations $(i)$ and $(ii)$ say that $$\frac{s}{t} < \exp\left(\frac{p}{q}\right) ;\ \ \frac{p}{q} < \frac{e^{ct/s}}{(s/t)^{t/s}} $$

Using equations $(iii)$ and $(iv)$ we conclude that $$ \exp\left(H_n\right) < \exp\left(\frac{p}{q}\right) ;\ \ \frac{p}{q} < \frac{e^{c\exp(-H_n)}}{(e^{H_n\!})^{\exp(-H_n)}}$$

Putting these together, we get that \begin{align} \exp(H_n) &< \exp\left(\frac{e^{c\exp(-H_n)}}{(e^{H_n\!})^{\exp(-H_n)}}\right)\\ \log(H_n) &< c\exp(-H_n) -H_n\exp(-H_n)\\ H_n+\log(H_n) \exp(H_n) &< c\\ \end{align} So by equation $(v)$ $$\sum_{d \mid n} d > H_n+\log(H_n) \exp(H_n)$$ which violates the form of the Riemann hypothesis in the linked MathOverflow question.

Conversely, if we had a violation of the Riemann hypothesis in that form, we could choose $m$ so that $s/t$ approximates $\exp(H_n)$ closely enough, and we could choose $k$ to approximate the other exponentials closely enough, and find a solution of these equations.

Yes. The Riemann Hypothesis ends up being equivalent to the following:

$$\forall n > 0, \left( \sum_{k \leq \delta(n)} \frac{1}{k} - \frac{n^2}{2} \right)^2 < 36n^3$$

This is proven using the same techniques from the proof of Hilbert's 10th Problem, and this is proven in the book Mathematical Developments Arising From Hilbert Problems with respect to the 10th problem.