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Source. Lemma 4, p. 5 of the arXiv PDF, proof pp. 5--6; it rests on Lemma 3 (p. 5). Read in the text layer and checked on the rendered pages.

Statement

Set $\delta_n=p_{n+1}-p_n$. Let $F\in\mathbb{Z}[x_0,\ldots,x_k]$ be a nonzero polynomial. Then, for almost all $n$,

$$F(\delta_n,\ldots,\delta_{n+k})\ne0 .$$

The paper does not define "almost all"; the proof bounds the exceptional $n$ in a range of length $x$ by $O(x/\log_2x)$, so the exceptions have density zero.

Proof (pp. 5--6), summarized

The proof works with indices $n$ of size about $x$, as its prime range $[p_x,p_{2x}]$ shows. Neglecting $O(x/\log_2x)$ indices ($\log_2$ the iterated logarithm), one may assume $\delta_i\le\log x\log\log x$ for $n\le i\le n+k$. For a fixed tuple $(\Delta_0,\ldots,\Delta_k)$ with $\Delta_i\le\log x\log_2x$, the number of $n$ with $\delta_{n+i}=\Delta_i$ for all $i$ is at most the number of primes $p\in[p_x,p_{2x}]$ such that $p+\Delta_0+\cdots+\Delta_i$ is prime for all $0\le i\le k$, which Lemma 3 bounds by $O(x\log_2^{k+2}x/\log^{k+1}x)$. Since $F\ne0$, the number of tuples in that box with $F(\Delta_0,\ldots,\Delta_k)=0$ is $O(\log^kx\log_2^kx)$. Multiplying, the number of these $n$ with $F(\delta_n,\ldots,\delta_{n+k})=0$ is $\ll x\log_2^{2k+2}x/\log x$, "which is sufficiently small" (p. 6). $\blacksquare$

Lemma 3, the sieve input, is stated on the Theorem 3 page with its citation to Halberstam–Richert; its proof is not in the paper.

Role

Used in the proof of Theorem 3 (pp. 8--9) to show that, after the recursion has removed all monomials $p_n^\nu/n^\mu$ with $\mu\ne\nu$, some coefficient polynomial $Q_i(\delta_n,\ldots,\delta_{n+\ell})$ is nonzero for almost all $n$; the same passage (p. 9) also uses that, for almost all $n$, none of $\delta_n,\ldots,\delta_{n+\ell}$ exceeds $\log^2n$.

Bears on. No catalog problem directly; it is a tool for Theorem 3, which is context for #251.