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Statement

Setting (pp. 1--2). $\Phi$ is the permutation of the 2-adic integers defined on the conjecture's page, and $C$ is the map $C(N)=N/2$ for even $N$, $C(N)=3N+1$ for odd $N$, on $\mathbb Z_2$ (see Theorem 1). $(1/3)\mathbb Z$ is the set of rationals $x$ with $3x\in\mathbb Z$, each a 2-adic integer.

Theorem 2 (p. 2): "If $C^k(N) = 1$ then $N \in \Phi((1/3)\mathbf{Z})$."

The print gives no range for $N$ or $k$; the proof applies to any $N\in\mathbb Z_2$ and any integer $k\ge0$. The paper presents the theorem as showing that the $3N+1$ conjecture (every positive integer has an iterate equal to $1$) implies its conjecture that $\mathbb Z^+\subseteq\Phi((1/3)\mathbb Z)$.

Proof pointer

Put $Q=\Phi^{-1}(N)$. By Theorem 1, $\Phi(H^k(Q))=C^k(N)=1$, so $H^k(Q)=\Phi^{-1}(1)=-1/3$ by display (3). The map $H$ pulls $(1/3)\mathbb Z$ back into itself (if $H(x)\in(1/3)\mathbb Z$ then $x\in(1/3)\mathbb Z$), and induction on $k$ gives $Q\in(1/3)\mathbb Z$ (p. 2).

Read depth

Claims checked: the statement was read on the page images of the print, and the proof was followed. A second reader checked the statement, hypotheses, label and page against the print; the proof was not independently reviewed.

Dependencies

Theorem 1 and display (3) (p. 1).

Source. Daniel J. Bernstein, A non-iterative 2-adic statement of the $3N+1$ conjecture, Proceedings of the American Mathematical Society 121 (1994), 405--408. Pages are those of the author's typescript named on the source card, numbered 1--4 rather than by the journal's pagination.

Bears on

• Problem 1135: one direction of the equivalence between the problem's question and the conjecture $\mathbb Z^+\subseteq\Phi((1/3)\mathbb Z)$; with Theorem 3 it makes that conjecture a restatement of the problem. It proves neither.