# Did Landau prove that there is a prime on $\bigl(x,\frac65x\bigr)$?

Was Landau the first to prove that there is a prime on $\bigl(x,\frac65x\bigr)$?

In his Handbuch $\!^1$ discussing the limit

$$\lim_{n\to\infty} \bigl(\pi\bigl((1+\epsilon)x\bigr)-\pi(x)\bigr)=\infty$$

he seems to say that in the next chapter he will prove the relation for all $\epsilon > \frac{1}{5}.$ On the face of it this is the answer and Jitsuro Nagura's proof$^2$ for $\epsilon \geq 1/5$ is not only an improvement but uses methods that Landau felt were exhausted by his proof, which is what I take from Landau's "da man doch nicht auf diesem elementaren Wege das Ziel erreichen kann..."

In the following section (21) of the following chapter (50) he derives a constant from sums involving the $\psi$ function and in section 22 proves Bertrand's Postulate. In 23 entitled "Weitere Verengerung der Schranken" [further narrowing of bounds] he performs another series of manipulations of $\psi(x)$ and derives that

$$(A)\hspace{7mm} \limsup_{x\to\infty}\frac{\psi(x)}{x}\leq\frac{171\cdot 6}{175\cdot 5}a \approx 1.08028$$ in which $a\approx 0.92129\dots$ and on the same page finds

$\psi(x)\geq ax+o(x) \approx 0.92129 x + o(x).$

So for comparison,

Nagura obtains

$0.916x-2.318 < \psi(x) < 1.086x$

and Landau obtains

$0.92129x + o(x) < \psi(x) < 1.08028 x.$

Landau halts his proof after (A), concluding that (A) "besser als $\frac{6a}{5}$ ist," leaving the reader I think a bit of work.

I haven't done the calculations yet but if as a quick check we substitute $0.93 x\leq \psi(x)\leq 1.085x$ into Nagura's expressions$^3$ for the difference $\vartheta\bigl(\frac{n+1}{n}x\bigr)-\vartheta(x)$ we get positive values for $n = 5$ and $x$ near $5000$.

So my question is whether Landau could have said $\epsilon\geq \frac{1}{5}$ using his own methods?

Landau and Nagura worked in different circumstances so this is really just curiosity about the nature of Landau's result.

$^1$ Handbuch (1909)p. 87.

$^2$ Nagura, On the Interval Containing At Least One Prime Number (1952).

$^3$ These values are a little worse than those Landau obtained assuming Landau's expression for the difference may involve some loss of numerical leverage over the expression in Nagura's paper.

• The answer is no. Landau's expression for the difference is the problem. – daniel Jan 11 '15 at 20:41
• And the proposition fails when $x=5.5$. – Oscar Lanzi Apr 10 '16 at 11:56

## 1 Answer

In § 20. Folgerungen über die Primzahlmenge zwischen $x$ und $(1+\varepsilon)x$, Landau proves that from an inequality

$$A \leqslant \liminf_{x\to\infty} \frac{\pi(x)\log x}{x} \leqslant \limsup_{x\to \infty} \frac{\pi(x)\log x}{x} \leqslant \varkappa A \tag{\ast}$$

the assertion

$$\lim_{x\to\infty} \bigl(\pi\bigl((1+\varepsilon)x\bigr) - \pi(x)\bigr) = \infty$$

follows for all $\varepsilon > \varkappa - 1$.

In § 22. Beweis des Bertrandschen Postulats, he derives $(\ast)$ with $\varkappa = \frac{6}{5}$ and $A = \log \frac{2^{1/2}3^{1/3}5^{1/5}}{30^{1/30}}$ in a sharper form, with explicit upper and lower bounds [for $\vartheta(x)$] that allow him to prove Bertrand's postulate. The bounds he has yield $\vartheta(2x-2) > \vartheta(x)$ for $x \geqslant 800$, and the remaining cases are covered by a list of small primes. Whether Landau is exactly following Chebyshev there or gives a slight modification of Chebyshev's proof I don't know, but that is of course irrelevant.

Then in § 23. Weitere Verengerung der Schranken, he first discusses the method used, and at the end reduces the factor $\varkappa$ in $(\ast)$ to $\frac{171}{175}\cdot \frac{6}{5}$, but he drops explicit bounds and just uses $o(x)$. So he proved

$$\lim_{x\to\infty} \bigl(\pi\bigl((1+\varepsilon)x\bigr) - \pi(x)\bigr) = \infty$$

for $\varepsilon = \frac{1}{5}$ - in fact, for all $\varepsilon > \frac{151}{875}$ - but he did not state an explicit $x_0$ such that $\pi\bigl(\frac{6}{5}x\bigr) - \pi(x) > 0$ for all $x \geqslant x_0$, let alone the smallest such $x_0$. He certainly could have given some such $x_0$, all that would have demanded is keeping explicit bounds. Whether these bounds would have yielded a small enough value $x_1$ that a list of "small" primes to cover the cases $25 \leqslant x \leqslant x_1$ could have been extracted from then-available prime tables is doubtful.

So indeed, the answer to "was Landau the first to prove that there is a prime in $\bigl(x,\frac{6}{5}x\bigr)$" is "no". If we interpret the statement as "for all large enough $x$", then Landau wasn't the first because that follows from the prime number theorem, proved (ever so slightly) before Landau's work. Also, that may already have been proved by Chebyshev or somebody in between extending Chebyshev's work. If we interpret the statement as "for all $x \geqslant 25$", then Landau did not prove that at all [probably; we cannot be sure that he never did].

but uses methods that Landau felt were exhausted by his proof, which is what I take from Landau's "da man doch nicht auf diesem elementaren Wege das Ziel erreichen kann …"

I think you're misinterpreting that. The full sentence is

Die Benutzung jener schärferen Kunstgriffe und jener längeren $U(x)$ führe ich hier nicht weiter aus, da man doch nicht auf diesem elementaren Wege das Ziel erreichen kann, die Abschätzungen der beiden Unbestimmtheitsgrenzen beliebig nahe aneinander zu bringen.

That is, the goal that cannot be reached using this elementary way is the prime number theorem. Landau was well aware that one could further improve the bounds using these methods, but since that wouldn't lead to the big goal, he wasn't particularly interested in pursuing them further. That he made the computation with explicit bounds to obtain a proof of Bertrand's postulate at all is probably only due to the fact that Bertrand's postulate was (and is) rather well-known.