Wiki
Wiki

Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

Updated


Statement

Setting. π(x)\pi(x) is the number of primes not exceeding xx. The paper recalls (p. 291, citing Landau's Handbuch, Vol. 1, §58) that π(2x)<2π(x)\pi(2x)<2\pi(x) for all sufficiently large xx.

Problem 1 (p. 291). Erdős asks whether

π(x+y)≤π(x)+π(y)(1)\pi(x+y)\le\pi(x)+\pi(y)\qquad(1)

holds, with no range on xx and yy stated. He records that Ungár has verified the inequality for y≤41y\le41, and that Hardy and Littlewood proved

π(x+y)−π(x)<cy/log⁡y(2)\pi(x+y)-\pi(x)<cy/\log y\qquad(2)

for a constant cc, deducing it by Brun's method.

The surrounding conjectures (p. 291). With ρ(y)=lim sup⁡x→∞ [π(x+y)−π(x)]\rho(y)=\limsup_{x\to\infty}\,[\pi(x+y)-\pi(x)], Hardy and Littlewood conjecture ρ(y)>y/log⁡y\rho(y)>y/\log y, and perhaps π(y)−ρ(y)→∞\pi(y)-\rho(y)\to\infty as y→∞y\to\infty. Erdős describes as very difficult, weaker than (1) and much stronger than (2), the conjecture that for each ε>0\varepsilon>0 there is yεy_\varepsilon such that for y>yεy>y_\varepsilon (quoted) "π(x+y)−π(y)<(1+ε)y/log⁡y\pi(x + y) - \pi(y) < (1 + \varepsilon)y/\log y" [sic]; the left side is printed with π(y)\pi(y) where the comparison with (1) and (2) suggests π(x)\pi(x). He adds that ρ(y)=1\rho(y)=1 for all yy has not been disproved, and draws a consequence for gaps between primes if ρ(y)>1\rho(y)>1 for some yy.

The paper poses (1) and does not resolve it.

Source. P. Erdős, Some unsolved problems, Michigan Math. J. 4 (1957), 291--300; §A, Problem 1, p. 291. The edition read is identified on the source card.

Read depth. Claims checked: the problem was read clause by clause on the page images of the journal print. A question has no proof to check; (2) is cited from Hardy and Littlewood, not proved here.

Dependencies

None.

Bears on

  • Problem 855: inequality (1) is the problem's inequality. The site's wording asks it for large xx and yy; the paper states no range. The paper poses the question and does not resolve it.