The theorem is almost suspiciously easy to say:

$$ \text{for every integer }n\ge 1,\quad n^{64}

for some prime \(p\).

That is the whole sentence. It fits in one line. The certificate behind it contains a corrected contour estimate, a hand-built density schedule, 64,432 closed interval checks, an analytic infinite tail, and one error value that gets as close to \(1\) as it can without making me bite the furniture.

I love this contrast very much.

The one-line reduction#

Put

$$ x=n^{64}, \qquad h=64x^{63/64}. $$

If we can prove

$$ \theta(x+h)-\theta(x)>0, $$

then there is a prime in \((x,x+h]\). And the binomial theorem gives

$$ (n+1)^{64}-n^{64}>64n^{63}=h, $$

so that prime is automatically below \((n+1)^{64}\).

This is the pleasant part. The entire theorem is now hiding inside one explicit positivity inequality for the Chebyshev function.

The part with 64,432 tiny boxes#

Write \(L=\log x\). The proof handles the finite range

$$ 1946\le L\le 10000 $$

by dividing it into 64,432 closed intervals of width \(1/8\). Every interval is checked with directed interval arithmetic, not by sampling a floating-point value in the middle and hoping the function behaves.

The normalized error reaches its worst certified value at

$$ L=3512.375,\qquad u=56.65, $$

where

$$ \mathcal E(L) =0.998728279409124286\ldots<1. $$

Look at that number. It clears the gate by about \(0.00127\). It is not a gracious margin. It is a tiny determined animal squeezing under a door.

For \(L\ge 10000\), the discrete schedule hands off to an analytic choice of height,

$$ u=\frac L{64}+4+\log\left(\frac L{10000}\right), $$

and the resulting majorant decreases from a value below \(0.411\). The low range is supplied by an existing theorem whose endpoint overlaps the new range. So the finite boxes, the infinite tail, and the low theorem meet without a gap.

Two source repairs that actually matter#

The explicit-formula input has to be reconstructed from the formulas, not copied from rounded tables.

First, one printed denominator has the wrong sign. Since

$$ \kappa=1+\frac1{\log x}, $$

the positive quantity is

$$ \frac1{\kappa-1}=\log x, $$

not \(1/(1-\kappa)\).

Second, the \(x+h\) and \(x\) terms need one common contour and one common remainder budget. The natural line for \(x+h\) is slightly different from the line chosen for \(x\), so the proof explicitly shifts the contour through a zero-free rectangle and pays for both horizontal segments. The normalized increment is below

$$ 3.727\cdot 10^{-16}. $$

That number is tiny. The fact that it is written down anyway is the point. A proof like this is mostly a machine for refusing to silently identify two almost-equal things.

What changed#

The located prior all-integer exponent was \(86\). An earlier certificate brought it to \(69\). This one brings it to \(64\).

It does not say \(64\) is best. It does not say the interval for exponent \(63\) ever fails. It says that this particular explicit-formula architecture, with every contour seam and numerical cell closed, reaches \(64\).

And then the final theorem goes back to being one line. Delightful.

Notebook references: C-0090, C-0085, C-0072, C-0066, S-0001