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Statement

Setting (p. 107). s1<s2<⋯s_1<s_2<\cdots are the squarefree numbers, and gt(x)g_t(x) is the number of si<xs_i<x with si+1−si=ts_{i+1}-s_i=t.

Lemma 1 (p. 107), cited as known from Mirsky (footnote 3): for fixed tt, as x→∞x\to\infty, gt(x)=βtx+o(x)g_t(x)=\beta_tx+o(x); that is, the density βt\beta_t of the sis_i with si+1−si=ts_{i+1}-s_i=t exists.

The moment asymptotic (23) (p. 107). Erdős says that, on hearing of Roth's bound (22), he thought of trying to prove for every α\alpha that

∑si+1≤x(si+1−si)α=Cαx+o(x).(23)\sum_{s_{i+1}\le x}(s_{i+1}-s_i)^\alpha=C_\alpha x+o(x).\qquad(23)

He says the proof of (23) seems very difficult, notes that it would imply si+1−si=o(siε)s_{i+1}-s_i=o(s_i^\varepsilon), and states that he can prove (23) only for α<A\alpha<A, where AA is a certain constant between 2 and 3. The paper sketches only the case α=2\alpha=2.

The case α = 2 (p. 109). The series ∑t≥1t2βt\sum_{t\ge1}t^2\beta_t converges, and

∑si+1≤x(si+1−si)2=x∑t=1∞t2βt+o(x).\sum_{s_{i+1}\le x}(s_{i+1}-s_i)^2=x\sum_{t=1}^{\infty}t^2\beta_t+o(x).

The paper says this "proves (23) for α=2\alpha=2". It gives no argument for other exponents, and the constant AA is not identified.

Source. P. Erdős, Some problems and results in elementary number theory, Publ. Math. Debrecen 2 (1951), 103--109, doi:10.5486/pmd.1951.2.2.04: Lemma 1 and (23) on p. 107, the sketch on pp. 108--109. Lemma 1 is cited there from L. Mirsky, Arithmetical pattern problems relating to divisibility by rr-th powers, Proc. London Math. Soc. 50 (1949), 497--508, Theorem 4, p. 507. The edition read is identified on the source card.

Read depth. Claims checked: the statements of Lemma 1, (23) and the case α=2\alpha=2 were read clause by clause on the printed pages, and the sketch was read through but not checked step by step. Lemma 1 was not checked against Mirsky's paper. Nothing here is independently reviewed.

Proof pointer

Pages 108--109. From Lemma 2, grouping the gaps into dyadic ranges (2r+j,2r+j+1](2^{r+j},2^{r+j+1}], for every ε>0\varepsilon>0 there is an rr with ∑si+1≤x, si+1−si>2r(si+1−si)2<εx\sum_{s_{i+1}\le x,\ s_{i+1}-s_i>2^r}(s_{i+1}-s_i)^2<\varepsilon x (28). From Lemma 1, the sum over gaps at most 2r2^r is x∑t≤2rt2βt+o(x)x\sum_{t\le2^r}t^2\beta_t+o(x) (29). Together these give a bound O(x)O(x) for the full sum, the convergence of ∑t2βt\sum t^2\beta_t, and the asymptotic.

Dependencies

Lemma 2 of the same paper, and Lemma 1 from Mirsky's paper cited above.

Bears on

  • Problem 145: the case α=2\alpha=2 shows that the limit the problem asks about exists for α=2\alpha=2, with value ∑t2βt\sum t^2\beta_t. The paper states, without proof, that it can reach every α<A\alpha<A with AA between 2 and 3. The range 0≤α≤20\le\alpha\le2 that the claim page Erdős 1951 records is the range later papers credit to this one; the paper prints only the case α=2\alpha=2.
  • Problem 489: the case A={p2:p prime}A=\{p^2:p\text{ prime}\}, whose BB is the squarefree numbers and which meets the problem's hypothesis ∣A∩[1,x]∣=o(x1/2)|A\cap[1,x]|=o(x^{1/2}); for it the mean squared gap tends to the finite limit ∑t2βt\sum t^2\beta_t. The paper's sum runs over si+1≤xs_{i+1}\le x where the problem's runs over bi<xb_i<x. It says nothing about any other AA beyond the remarks on p. 109.