In this post I will sketch a possible proof of the Riemann hypothesis. In 2002, it was proved that the Riemann hypothesis is equivalent to the inequality

\[\sigma(n) \leq H_n + \ln (H_n) e^{H_n}\]

where \(\sigma(n)\) is the divisor function (the number of divisors of \(n\)) and \(H_n\) is the harmonic series \(1 + 1/2 + 1/3 + \ldots + 1/n\). By showing that the inequality is true, it holds that the Riemann hypothesis is also true.

Lemma 1

Statement: \(\sigma(n) \leq n\).

Proof: If \(d\) is a divisor of \(n\), then \(1 \leq d \leq n\) and only \(n\) distinct values for \(d\) are possible.

Lemma 2

Statement: \(n \uparrow\uparrow 3 \leq n \uparrow\uparrow\uparrow 2\) for \(n \geq 3\).

Proof: By induction:

\[n \uparrow\uparrow 3 \leq n \uparrow\uparrow 2\]

\[e^{n \uparrow\uparrow 3} \leq e^{n \uparrow\uparrow 2}\]

\[\arctan(e^{n \uparrow\uparrow 3}) \leq \arctan(e^{n \uparrow\uparrow 2})\]

\[\arctan(e^{n \uparrow\uparrow 3})^{\cos \int_1^\infty \frac{dx}{\ln x}} \leq \arctan(e^{n \uparrow\uparrow 2})^{\cos \int_1^\infty \frac{dx}{\ln x}}\]

\[e^{(n + 1) \uparrow\uparrow 3} \leq e^{(n + 1) \uparrow\uparrow 2}\]

And we have the base case \(\pi = \sqrt{pi^2}\).

Lemma \(3.0168\)

I conjecture that the exact value of the lemma is \(\sqrt{\sqrt[5]{\phi} + 8}\). This makes intuitive sense — the Axiom of Choice tells us that all complex analysis proofs must have an irrational-numbered lemma.

Statement: \(e^\pi - \pi = 20\).

Proof: By contradiction. If the statement is true, then by the 1=0 theorem (see Appendix B) it is false. Likewise, if it is false, it is true. This creates a paradox, so the only possibility is that it is true. Note that since \(\omega^{\omega^\omega} = 20\), which implies that connecting points A, B, and D produces an isosceles triangle.