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Statement

Theorem III (p. 609). Let

$$a_1<a_2<\cdots<a_{k+1},\qquad b_1<b_2<\cdots<b_v$$

be the two sets, which the paper introduces as sets of positive integers (p. 609). The sums $a_i+b_j$ cannot all be composed of only $k$ primes if one of the $b$'s is greater than $a_{k+1}^k$. The paper adds that this surely occurs if $v>a_{k+1}^k$; it gives no reason, and the reason is that then $b_v\ge v$ for increasing positive integers (an observation of this page).

Before the theorem (p. 609) the paper poses the question whether two infinite sets $a_1<a_2<\cdots$ and $b_1<b_2<\cdots$ of positive integers can have every sum $a_i+b_j$ composed of given primes $p_1,\ldots,p_k$, and answers it in the negative; Theorem III is the stronger finite form, since the first $k+1$ of the $a$'s and a $b$ above $a_{k+1}^k$ already give a contradiction.

Source. Paul Erdős and Paul Turán, On a problem in the elementary theory of numbers, Amer. Math. Monthly 41 (1934), 608-611: the question and Theorem III on p. 609, the proof in Section 5 on p. 611. The edition read is identified on the source card.

Read depth. Claims checked: the statement and the proof were read clause by clause on the printed pages. Nothing here is independently reviewed.

Proof pointer

Section 5, p. 611. Assume $b_v>a_{k+1}^k$ and all sums composed of $p_1,\ldots,p_k$. Each of the $k+1$ sums $a_l+b_v$ exceeds $a_{k+1}^k$ and has at most $k$ prime factors, so some prime power $p^{\alpha}$ exactly dividing it exceeds $a_{k+1}$; call $p$ the prime belonging to $a_l$. If the same prime belonged to two of the $a$'s, the smaller of the two prime powers would divide their difference, a positive integer below $a_{k+1}$, while exceeding $a_{k+1}$. So the $k+1$ integers $a_l$ have distinct primes, which $k$ primes cannot supply.

Dependencies

None beyond elementary divisibility.

Bears on

None of the corpus's problem pages.