MATHEMATICS

P. ERDŐS (P. ERDŐS), A. SÁRKÖZI (A. SARKÖZI),
E. SZEMERÉDI (E. SZEMERÉDI)

ON THE SOLVABILITY OF CERTAIN EQUATIONS IN DENSE SEQUENCES OF INTEGERS

(Presented by Academician A. N. Kolmogorov, 4 IV 1967)

In the preceding paper \((^1)\), using a simple combinatorial result of Kleitman \((^4)\), we showed that if \(a_1<a_2<\cdots\) is an infinite sequence of integers such that for infinitely many \(x\) the inequality

\[ Ax=\sum_{a_i\leq x}\frac{1}{a_i}>c_1\log x/(\log\log x)^{1/2}, \]

holds, then the equations \((a_i,a_j)=a_r,\ r<i<j,\ [a_{i_1},a_{j_1}]=a_{r_1},\ i_1<j_1<r_1,\) have infinitely many solutions. We also showed that, in a certain sense, this theorem cannot be improved, namely, that the constant \(c_1\) cannot be replaced by an arbitrarily small constant. More precisely, we constructed a sequence satisfying the condition

\[ \sum_{a_i\leq x}1>c_2x/(\log\log x)^{1/2}, \tag{1} \]

but nevertheless the equation \([a_{i_1},a_{j_1}]=a_{r_1},\ i_1<j_1<r_1,\) is not solvable.

In this paper, by \(c,c_1,c_2,\ldots\) we shall denote absolute constants; by \(p\), prime numbers; by \(P(n)\), the largest, and by \(p(n)\), the smallest prime divisor of the number \(n\). The sequence \(a_1<a_2<\cdots\) will be denoted by \(A\).

We shall say that a sequence \(u_1<u_2<\cdots\) satisfies property I if the equation \(u_iq=u_j,\ p(q)>P(u_i)\) has no solutions.

In the present paper we shall show that the equation \((a_i,a_j)=a_r\) behaves quite differently from the equation \([a_i,a_j]=a_r\). We shall prove the following theorem:

Theorem. Let \(a_1<\cdots\) be a sequence of integers for which the equation

\[ (a_i,a_j)=a_r,\quad r<i<j, \tag{2} \]

has no solutions. Then

\[ \sum \frac{1}{a_i\log a_i}<c. \tag{3} \]

We shall make some preliminary remarks. By partial summation, from the theorem of our paper \((^2)\) we easily obtain that, if equation (2) has no solutions, then for every \(k\) the equality

\[ \liminf_{x\to\infty}\sum_{a_i\leq x} \left( \frac{x}{\prod_{r=2}^{k}\log_r x} \right)^{-1}=0 \]

holds (by \(\log_r x\) is denoted the \(r\)-th iteration of the logarithm).

Thus, relations similar to (1) cannot occur in this case.

A sequence \(b_1<\cdots\) is called primitive if there does not exist a number \(b\) by which all the remaining terms of the sequence are divisible.

... It is well known (³) that for every primitive sequence the inequality

\[ \sum \frac{1}{b_i\log b_i}<c_3, \tag{4} \]

holds, and also (see (2)) the equality

\[ \lim_{x=\infty}\sum_{b_i\le x}\frac{1}{b_i} \left(\frac{\log x}{(\log\log x)^{1/2}}\right)^{-1}=0, \tag{5} \]

holds; moreover, this relation cannot be improved.

We prove that if \(a_1<a_2<\cdots\) is an infinite sequence for which equation (2) is not solvable, then

\[ \lim_{x\to\infty}\sum_{a_i\le x}\frac{1}{a_i} \left(\frac{\log x}{(\log\log x)^{1/2}}\right)^{-1}=0. \tag{6} \]

The proof of equality (6) is quite complicated, and we shall return to it later. Relations (3), (4), (5), and (6) give rise to the following question. Let \(b_1<b_2<\cdots\) be an infinite primitive sequence. Does there exist a constant \(c>0\) and a sequence \(a_1<\cdots\) for which equation (2) is not solvable and \(a_n\le b_n^c\)? We cannot answer this question.

We now pass to the proof of the theorem. We use the following lemma, due to Alexander.

Lemma 1. Let \(a_1<a_2<\cdots\) be a sequence satisfying property I. Then

\[ \sum_i \frac{1}{u_i\log u_i}<c_4. \tag{7} \]

If \(u_i\nmid u_j\) (i.e., if the sequence \(u_1<u_2<\cdots\) is primitive), then inequality (7) was proved in (³). The proof of Lemma 1 is similar to the proof given in (³), but for completeness we shall give a sketch of it here. It is not hard to see that condition I means (see (³)) that \(u_iq=u_jq'\), \(p(q)>P(u_i)\), \(p(q')>P(u_j)\).

Using the sieve of Eratosthenes, we conclude that the number of integers \(u_iq\le x\), \(p(q)>P(u_i)\), is greater than

\[ \prod_{p\le P(u_i)}\left(1-\frac{1}{p}\right)-2^{u_i}. \tag{8} \]

From relation (8) we easily obtain the inequality

\[ \sum_i \prod_{p\le P(u_i)}\left(1-\frac{1}{p}\right)/u_i\le 1, \tag{9} \]

from which, with the aid of Mertens’ theorem

\[ \prod_{p<y}\left(1-\frac{1}{p}\right)<c/\log y, \]

the proof of our lemma follows.

We now define a subsequence \(A(a_i)\) of the sequence \(A\) as follows: \(a_j\) is contained in \(A(a_i)\) if the number \(a_i\) is the greatest of those \(a\) for which the equation \(a_j=a_iq,\ p(q)>P(a_i)\), is solvable. Let \(A'\) be the subsequence of the sequence \(A\) which is not contained in any subsequence \(A(a_i)\). It is clear that

\[ A=A'\cup \bigcup_{i=1}^{\infty} A(a_i). \]

Thus,

\[ \sum_k \frac{1}{a_k\log a_k} = \sum_{a_k\in A'}\frac{1}{a_k\log a_k} + \sum_{i=1}^{\infty}\sum_{a_k\in A(a_i)} \frac{1}{a_k\log a_k}. \tag{10} \]

Obviously, the subsequence \(A'\) satisfies property I. Thus, by Lemma 1, the inequality

\[ \sum_{a_k\in A'} \frac{1}{a_k\log a_k}<c_4 . \tag{11} \]

holds.

We now prove Lemma 2.

Lemma 2.

\[ \sum_{a_k\in A(a_i)} \frac{1}{a_k\log a_k}<\frac{c_5}{a_i P(a_i)^{1/2}} . \]

It is easy to see (the \(q_1<q_2<\cdots\) run through the set of all primes) that

\[ =\sum \frac{1}{n(P(n))^{1/2}} =\sum_{m=1}^{\infty}\frac{1}{q_m^{3/2}}\prod_{i=1}^{m}\left(1+\frac{1}{q_i}\right) <\sum_{m=1}^{\infty}\frac{c\log q_m}{q_m^{3/2}}<\infty . \]

Our Theorem 1 therefore follows immediately from (10), (11), and Lemma 2. To complete the proof it remains only to establish Lemma 2. Let \(a_iq_r^{(i)}\), \(r=1,\ldots\), \(p(q_r^{(i)})>P(a_i)\), be the integers of the subsequence \(A(a_i)\). It is clear that the sequence \(q_r^{(i)}\) satisfies property I. If this is not so and \(q_{r_2}^{(i)}/q_{r_1}^{(i)}\) is an integer satisfying the inequality \(p(q_{r_2}^{(i)}/q_{r_1}^{(i)})>P(q_{r_1}^{(i)})\), then \(a_iq_{r_2}^{(i)}\) (which belongs to the subsequence \(A(a_i)\)) can be written in the form \(a_lq\), \(p(q)>P(a_l)\), \(a_l=a_iq_r^{(i)}\), \(q_{r_2}^{(i)}/q_{r_1}^{(i)}=q\), which contradicts the maximality of \(a_i\).

We now show that there do not exist two relatively prime numbers \(q_r^{(i)}\). To see this, we first use the fact that equation (2) has no solutions. Namely, assuming that \((q_{r_1}^{(i)},q_{r_2}^{(i)})=1\), we obtain \((a_iq_{r_1}^{(i)},a_iq_{r_2}^{(i)})=a_i\). In other words, equation (2) has a solution, contrary to our assumption.

Lemma 3. Let the sequence \(q_1<\cdots\) satisfy property I, \((q_i,q_j)\ne 1\), and \(p(q_i)>t\). Then

\[ \sum_i \frac{1}{q_i\log q_i}\ll \frac{c_5}{t^{1/2}} . \]

Lemma 2 immediately follows from Lemma 3, since

\[ \sum_{a_k\in A(a_i)}\frac{1}{a_k\log a_k} =\sum_r \frac{1}{a_i q_r^{(i)}\log a_iq_r^{(i)}} \ll \frac{1}{a_i}\sum_r \frac{1}{q_r^{(i)}\log q_r^{(i)}} <\frac{c_5}{a_i p(a_i)^{1/2}} . \]

Thus, it remains only for us to prove Lemma 3. It is quite likely that Lemma 3 is not the best possible, and that the expression \(c_5/t^{1/2}\) can be replaced by \(c_6/t\).

For the proof of Lemma 3 we first assume that there exists an \(i\) for which

\[ \sum_{p/q_i}\frac{1}{p}\ll \frac{1}{t^{1/2}} . \tag{12} \]

Since there do not exist two relatively prime numbers \(q_r\), every \(q^r\) must be divisible by at least some \(p\), where \(p\mid q_i\). Therefore

\[ \sum_r \frac{1}{q_r\log q_r} \ll \sum_{p/q_i}\frac{1}{p}\sum' \frac{1}{q_r/p\log q_r}, \tag{13} \]

where the prime indicates that the summation extends over all \(q\) such that \(p\mid q\). Obviously, the sequence \(q_r/p\) satisfies property I (except that one of the numbers \(q_r/p\) may be equal to one).

Thus, by Lemma 1,

\[ \sum' \frac{1}{q_r \log q_r} < 1+c_3. \tag{14} \]

From inequalities (12), (13), and (14) we obtain that

\[ \sum_r \frac{1}{q_r \log q_r} < (1+c_3)\sum_{p/q_i}\frac{1}{p} \leq \frac{1+c_3}{t^{1/2}} . \]

This proves the lemma.

To complete our proof, suppose now that inequality (12) is false for \(q_r\). Let \(l\) be an integer and let \(x>x_0(l)\) be large. Consider the integers not exceeding \(x\) of the form \(q_r(t)\), where all prime factors of \(t\) are greater than \(q_r\). Since the sequence \(q_r\) satisfies property I, we obtain, as in Lemma 1, that the integers

\[ q_r m,\quad r=1,2,\ldots,l,\quad m<x/q_r, \tag{15} \]

are distinct. Denote by \(u_1,u_2,\ldots,u_s\) the numbers of the form (15). By Mertens’ theorem and the sieve of Eratosthenes, we obtain

\[ s=(1+O(1))\sum_{r=1}^{l}\frac{x}{q_r}\prod_{p=P(q_r)}\left(1-\frac{1}{p}\right) > Cx\left(\sum_r \frac{1}{q_r \log q_r}\right)+O(x). \tag{16} \]

Obviously, all prime factors of \(u\) are greater than \(t\), and since inequality (12) is false, we have

\[ \sum_{p/u_i}\frac{1}{p}>\frac{1}{t^{1/2}} . \]

Thus, on the one hand,

\[ \sum_{i=1}^{s}\sum_{p/u_i}\frac{1}{p}>\frac{s}{t^{1/2}}, \tag{17} \]

and on the other,

\[ \sum_{i=1}^{s}\sum_{p/u_i}\frac{1}{p} < \sum_{u=1}^{x}\sum_{\substack{p/u\\ p>t}}\frac{1}{p} < \sum_{p>t}\frac{x}{p^2} < \frac{x}{t}. \tag{18} \]

Therefore, from inequalities (17) and (18) we obtain the inequality

\[ s < x/t^{1/2}. \tag{19} \]

Thus inequalities (16) and (19) give the inequality

\[ \sum_{r=1}^{l}\frac{1}{q_r \log q_r}<c_5 t^{1/2}, \tag{20} \]

and since the last inequality is valid for every \(l\), the proof of Lemma 3, and consequently of the theorem, is complete.

Our proof does not use Kleitman’s combinatorial result \({}^{4}\). We do not know how to deal with the equation \([a_i,a_j]=a_r\) without using Kleitman’s result.

Mathematical Institute
Hungarian Academy of Sciences
Budapest, Hungary

Received
4 IV 1967

CITED LITERATURE

\({}^{1}\) P. Erdös, A. Sárközi, E. Szemerédi, J. Math. Anal. and Appl., 15, 60 (1966).
\({}^{2}\) P. Erdös, A. Sárközi, E. Szererédi, J. Australian Math. Soc., 7, 9 (1967).
\({}^{3}\) P. Erdös, London Math. Soc. J., 10, 116 (1935).
\({}^{4}\) D. Kleitman, Proc. Am. Math. Soc., 17, 139 (1966).