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Source. Lemma 1, pp. 229--230, of Paul Erdős, Carl Pomerance and András Sárközy, On locally repeated values of certain arithmetic functions, IV, The Ramanujan Journal 1 (1997), 227--241, DOI 10.1023/A:1009723712317, as identified on the source card.

Statement

Lemma 1 (pp. 229--230). Let x≥1x\ge1, m∈Nm\in\mathbb N with

m≤x1/2,(2.1)m\le x^{1/2},\qquad(2.1)

h∈Zh\in\mathbb Z, and let ff be a non-negative additive arithmetic function with

f(pα)=0for p∣m, α∈N.(2.2)f(p^\alpha)=0\quad\text{for } p\mid m,\ \alpha\in\mathbb N.\qquad(2.2)

Put

K=max⁡{f(pα):pα≤x},A=∑p≤xf(p)p.(2.3)K=\max\{f(p^\alpha):p^\alpha\le x\},\qquad A=\sum_{p\le x}\frac{f(p)}{p}. \qquad(2.3)

Then

∑n≤xn≡h (mod m)(f(n)−A)2<c3 xm (KA+K2),\sum_{\substack{n\le x\\ n\equiv h\ (\mathrm{mod}\ m)}}(f(n)-A)^2 <c_3\,\frac xm\,(KA+K^2),

where c3c_3 is an absolute constant, independent of xx, mm, hh and ff.

Remarks the paper makes without proof (p. 230). The hypothesis (2.1) may be replaced by m≤x1−δm\le x^{1-\delta} for any fixed δ\delta with 0<δ<10<\delta<1, with c3c_3 then depending on δ\delta. For completely additive ff, KK may be taken as the maximum of f(p)f(p) over primes p≤xp\le x, at the cost of a larger absolute constant. The sign condition can be removed by splitting a real additive function into non-negative and non-positive parts, and a complex one into real and imaginary parts, with f(pα)f(p^\alpha) replaced by ∣f(pα)∣\lvert f(p^\alpha)\rvert in the definition of KK. The paper contrasts the lemma with earlier inequalities of this kind (Alladi; Kubilius), whose moduli must be much smaller, fixed or at most a power of log⁡log⁡x\log\log x: the quantity KK in the bound is what allows moduli as large as a power of xx.

Read depth. Claims checked: the statement and the remarks were read clause by clause on the printed pages. The proof (pp. 230--232) was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

Pages 230--232. The proof truncates ff to f1f_1, which keeps f(pα)f(p^\alpha) for pα≤x1/4p^\alpha\le x^{1/4} and is zero above. Since n≤xn\le x has at most three exactly dividing prime powers above x1/4x^{1/4}, ∣f(n)−f1(n)∣≤3K\lvert f(n)-f_1(n)\rvert\le3K, and the mean A1A_1 of f1f_1 differs from AA by O(K)O(K). The first and second moments of f1f_1 over the progression are then computed by counting the n≤xn\le x in the class exactly divisible by pαp^\alpha, or by both pαp^\alpha and qβq^\beta, which (2.1) and (2.2) make possible with errors of size at most O(K2x1/2)O(K^2x^{1/2}). Expanding the square gives the bound for f1f_1, and the truncation estimates transfer it to ff.

Dependencies

None beyond elementary prime sums; the paper cites no earlier result in the proof.

Used in the paper

Lemma 2 (p. 233) applies Lemma 1 to ωm(n)\omega_m(n), the number of primes dividing nn but not mm (so K=1K=1 and A=log⁡log⁡x+O(log⁡log⁡log⁡x)A=\log\log x+O(\log\log\log x)), and obtains absolute constants c4,x0c_4,x_0 such that, for x>x0x>x_0, m≤x1/2m\le x^{1/2} and h∈Zh\in\mathbb Z, more than 12x/m\tfrac12x/m of the n≤xn\le x with n≡h(modm)n\equiv h\pmod m have $\lvert\omega_m(n)-\log\log x\rvert<c_4(\log\log x)^{1/2}$. This is the input to the proof of Theorem 1.

Bears on

Lemma 1 bears on no Erdős problem directly. It is the main input to Theorem 1, whose page states that theorem's relation to Problem 122.