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Statement

Notation as on the Theorem I page: $P_i$ is the $i$-th prime and (1) is $\pi(x+y)\leq\pi(x)+\pi(y)$.

Lemma IV (p. 525; quoted). "(1) is false for some integers $x\geq2$, $y\geq2$, if and only if there exists a prime number, $P_n$, and an integer $q$, $1\leq q\leq(n-1)/2$, such that

$$P_{n-q}+P_{q+1}-3\geq P_n\geq P_{n-q}+P_q+1."$$

The double inequality is the paper's display (9). The "if" direction is constructive (p. 525): when (9) holds, the pair

$$x=P_n-P_{n-q}+1,\qquad y=P_{n-q}-1$$

has $x+y=P_n$ and $\pi(x+y)>\pi(x)+\pi(y)$. By (9) this $x$ lies between $P_q+2$ and $P_{q+1}-2$, so it is large only when $q$ is.

Source. Sanford L. Segal, On $\pi(x+y)\leq\pi(x)+\pi(y)$, Trans. Amer. Math. Soc. 104 (1962), no. 3, 523--527, doi:10.1090/s0002-9947-1962-0139586-4: Lemma IV and its proof on p. 525, using Lemmas I--III on pp. 523--525. The edition read is identified on the source card.

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

Proof pointer

P. 525. "If": for the pair above, $\pi(x+y)=n$ and $\pi(y)=n-q-1$, while the upper bound in (9) gives $x\leq P_{q+1}-2$, so $\pi(x)\leq q$. "Only if": the paper's Lemmas I--III give integers $M_0,K_0\geq2$ with $\pi(M_0+K_0)=\pi(M_0)+\pi(K_0)$, $M_0+K_0+1$ and $K_0+1$ prime, $M_0+1$ composite and odd, and $K_0\geq M_0+2$. Writing $M_0+K_0+1=P_n$ and $K_0+1=P_{n-q}$ makes $q=\pi(M_0)$, so $P_q+1\leq M_0\leq P_{q+1}-3$, and adding $P_{n-q}$ gives (9). The bound $K_0\geq M_0+2$ gives $P_{n-q}\geq P_q+4$, hence $q\leq(n-1)/2$.

Dependencies

• Lemma I (pp. 523--524): (1) fails for some $x,y\geq2$ exactly when some integers $M,K\geq2$ have $\pi(M+K)=\pi(M)+\pi(K)$, $M+K+1$ prime and $M+1$ composite.
• Lemma II (p. 524): with $M_0$ the least such $M$, a corresponding $K_0$ has $K_0+1$ prime.
• Lemma III (pp. 524--525): for these $M_0,K_0$, $K_0\geq M_0+2$.

Bears on

• Problem 855: the lemma turns a violation of the inequality into a prime in the window (9), and back. A family of solutions of (9) along which both $P_n-P_{n-q}+1$ and $P_{n-q}-1$ tend to infinity would give violations with both variables large and so answer the problem negatively; since $x$ stays below $P_{q+1}$, this needs $q\to\infty$. The paper exhibits no solution of (9).