Can one prove by elementary means (such as Paul Erdös' proof of Bertrand's Postulate) that every prime greater than 5 is less than the sum of the two primes immediately preceding it?
$\begingroup$
$\endgroup$
3
asked Oct 10, 2018 at 18:59
-
5$\begingroup$ From this, $p_{n+2}\leq \frac{3}{2}p_{n+1}\leq p_{n+1}+\frac{3}{4}p_n$. $\endgroup$– WojowuCommented Oct 10, 2018 at 19:11
-
3$\begingroup$ Prof. Bernardo Recamán Santos: the answer is YES, essentially this question was the subject matter of this previous question in MO: mathoverflow.net/q/113840/1593 $\endgroup$– José Hdz. Stgo.Commented Oct 10, 2018 at 19:13
-
3$\begingroup$ @Wojowu: El Bachraoui's result was already established by Schur in 1929, see Jose Brox's answer here: mathoverflow.net/questions/289402/… . $\endgroup$– Ofir GorodetskyCommented Oct 10, 2018 at 19:16
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
I am to elaborate a bit on how it is that the answer I left in this previous question implies a positive answer to Mr. Recamán's question.
By elementary means (i.e., without resorting to any complex analysis), in his paper Mémoire sur les nombres premiers (Mémoires de l'Acad. Imp. Sci. de St. Pétersbourg, VII, 1850.), P. L. Chebyshev established that for any $\varepsilon >\frac{1}{5}$, there exists an $n_{\varepsilon}\in \mathbb{N}$ such that for all $n\geq n_{\varepsilon}$,
$$\pi((1+\varepsilon)n)−\pi(n)>0.$$
In particular, taking $\varepsilon = \frac{1}{4}$, this implies that, for every sufficiently large $k \in \mathbb{N}$, we have that
$$p_{k+2} \leq (1.25)^{2}p_{k} < 2p_{k}$$
and whence, for every sufficiently large $k \in \mathbb{N}$, it is certainly the case that
$$p_{k+2} < 2p_{k} < p_{k}+p_{k+1}.$$
By retracing Chebyshev's arguments it is possible to obtain an explicit version of the previous conclusion.
answered Oct 10, 2018 at 19:43