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Statement

Lemma II (p. 104). Let $n\ge3$ and let $w\equiv1\pmod{16}$. For $0\le i\le n-1$ let $B_i$ be a divisor of the Fermat number $2^{2^i}+1$ with $B_i>1$; $B_i$ need be neither prime nor smaller than $2^{2^i}+1$. Suppose

$$w\prod_{i=0}^{n-1}B_i\le2^{2^n}-1 .$$

Then $w\prod_{i=0}^{n-1}B_i$ is not of the form $p+2^a+2^b$ with $p$ prime and $a,b$ distinct positive integers.

The paper's conventions (p. 103) apply: all quantities are integers, usually positive, and a prime is a positive prime. The exponents are positive and distinct, so neither the exponent-zero case nor two equal powers is covered by the lemma.

Footnote 2 (p. 104) remarks that the modulus $16$ for $w$ is not essential: any power of $2$ larger than $16$ would serve, for example $w\equiv1$ or $w\equiv3\pmod{64}$, with only trivial changes in what follows.

Source. R. Crocker, On the sum of a prime and of two powers of two, Pacific J. Math. 36 (1971), no. 1, 103-107; Lemma II on p. 104, its proof on pp. 104-105. The copy read is identified on the source card.

Read depth. Claims checked: the statement was read clause by clause on the page images, and the proof was read; its residue computation is summarized below but was not independently reviewed.

Proof pointer

Pages 104-105. The lemma generalizes Lemma I (p. 104), which is the case $w=1$, $B_i=2^{2^i}+1$: for $n\ge3$, $2^{2^n}-1$ is not a prime plus two distinct positive powers of $2$. For $n=3,4,5$ the paper says Lemma I applies directly. (The reason, not spelled out in the paper: $2^{2^i}+1$ is prime for $i\le4$, so each $B_i$ is the whole Fermat number, the product of the $B_i$ is $2^{2^n}-1$, and the size bound forces $w=1$.) For $n\ge6$, take $a>b$; both are below $2^n$, and with $2^r$ the exact power of $2$ dividing $a-b$, the factor $2^{2^r}+1$ divides $2^{a-b}+1$, so $B_r$ divides $w\prod B_i-2^b(2^{a-b}+1)$, which is positive. It remains to rule out that this difference equals $B_r$. The paper splits the product of the $B_i$ at $i=4$ and, recalling $B_i\equiv1\pmod{2^{i+1}}$, so $B_i\equiv1\pmod{16}$ for $i\ge5$, finds it $-1$ modulo $16$; hence so is $w\prod B_i$. The product exceeds $B_0B_1B_r=15B_r$, so equality would give $2^a+2^b>14B_r\ge42$, forcing $a>3$; then modulo $16$ the sum $2^a+2^b+B_r$ is $0$ plus one of $0,2,4,8$ plus one of $1,3,5$, never $-1$. So the difference is a proper multiple of $B_r$ and is not prime.

Dependencies

Lemma I of the same paper (p. 104), which the author says was communicated to him by A. Schinzel and which also appears in Sierpiński's Elementary Theory of Numbers (footnote 1); both lemmas come from the method of the author's 1960/61 note in Mathematics Magazine (the paper's reference [1]).

Bears on

• Problem 9: the lemma is the step of Theorem I that excludes a prime plus two distinct positive powers of $2$; on its own it says nothing about density.
• Problem 10: through the construction of Theorem I, the lemma gives the constructed integers the exclusion of a prime plus two distinct positive powers of $2$ that the parity argument for the settled Grechuk variant uses; it does not bear on the problem's main question.