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

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Statement

Theorem 3 (p. 113, quoted). "There exists an arithmetic progression consisting only of odd numbers, no term of which is of the form 2k+p2^k+p."

Here pp denotes a prime. The paper proves it following a question of Romanoff, communicated in writing (p. 113, footnote 3).

The construction (p. 119). Every integer kk lies in at least one of the classes

0 (mod 2),0 (mod 3),1 (mod 4),3 (mod 8),7 (mod 12),23 (mod 24),0\ (\mathrm{mod}\ 2),\quad 0\ (\mathrm{mod}\ 3),\quad 1\ (\mathrm{mod}\ 4),\quad 3\ (\mathrm{mod}\ 8),\quad 7\ (\mathrm{mod}\ 12),\quad 23\ (\mathrm{mod}\ 24),

and the paper takes xx in the classes it prints as 1(mod2)1\pmod2, 1(mod7)1\pmod7, 2(mod5)2\pmod5, 23(mod17)2^3\pmod{17}, 27(mod13)2^7\pmod{13} and 223(mod241)2^{23}\pmod{241}, so that for every kk the number x−2kx-2^k is a multiple of one of the primes 3,5,7,13,17,2413,5,7,13,17,241. The six classes of kk pair in order with the primes 3,7,5,17,13,2413,7,5,17,13,241, since the order of 22 modulo these primes is 2,3,4,8,12,242,3,4,8,12,24. The class for the prime 33 is not printed: the pairing of 0(mod2)0\pmod2 with 33 requires x≡1(mod3)x\equiv1\pmod3, which the corpus reads as intended alongside the printed x≡1(mod2)x\equiv1\pmod2 that makes xx odd. The paper does not discuss a term xx for which x−2kx-2^k equals one of the six primes itself.

Remarks after the proof (p. 120). The paper explains that the method works because, for n≠6n\ne6, some prime divides 2n−12^n-1 but no 2m−12^m-1 with m<nm<n (its footnote 9, Landau's tract), and that the simplest covering system with distinct moduli, 0(mod2)0\pmod2, 0(mod3)0\pmod3, 1(mod4)1\pmod4, 5(mod6)5\pmod6, 7(mod12)7\pmod{12}, cannot be used because of the modulus 66. It lists a covering system without the modulus 22, with moduli 3,4,5,6,8,10,12,15,20,24,30,40,60,1203,4,5,6,8,10,12,15,20,24,30,40,60,120, and records that Davenport found a slightly more complicated system earlier. The conjecture that follows is on its own page.

Proof pointer

P. 119, the construction above: a residue class modulo 2⋅3⋅5⋅7⋅13⋅17⋅2412\cdot3\cdot5\cdot7\cdot13\cdot17\cdot241 chosen by the Chinese remainder theorem.

Read depth

Claims checked: the statement on p. 113, the proof on p. 119 and the remarks on p. 120 were read on the page images of the print; the congruences were checked against the orders of 22 named above. Nothing here is independently reviewed.

Dependencies

None in the corpus. External input named by the paper: the existence of a primitive prime divisor of 2n−12^n-1 for n≠6n\ne6, cited through Landau's tract (its footnote 9).

Source. P. Erdős, On integers of the form 2k+p2^k+p and some related problems, Summa Brasil. Math. 2 (1950), fasc. 8, 113--123; the edition read is named on the source card.

Bears on

  • Problem 16: the theorem supplies an infinite arithmetic progression inside the set of odd integers not of the form 2k+p2^k+p, the progression part of the decomposition the problem asks about; it says nothing about whether the rest of that set has density 00. The problem's claim page records the later work on the question.