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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Statement

In the note a pseudoprime is a composite natural number nn with n∣2n−2n\mid 2^n-2 (p. 816).

Théorème 1 (p. 816, quoted). "Le plus petit nombre entier k>1k>1 pour lequel il existe un nombre pseudopremier nn tel que le nombre knkn est aussi pseudopremier est le nombre k=23k=23."

That is: for no integer kk with 1<k<231<k<23 is there a pseudoprime nn with knkn a pseudoprime, and for k=23k=23 there is one. The proof (p. 817) names the pair n=89⋅683=60787n=89\cdot683=60787 and kn=23⋅89⋅683=1398101kn=23\cdot89\cdot683=1398101.

Lemme 1 (p. 816), the reduction the proof rests on: if nn and knkn are both pseudoprimes, then n∣2k−2n\mid 2^k-2. The note proves it for odd nn and for even n=2n1n=2n_1 separately.

Source. A. Rotkiewicz, Sur les nombres naturels n et k tels que les nombres n et nk sont à la fois pseudopremiers, Atti Accad. Naz. Lincei Rend. Cl. Sci. Fis. Mat. Nat. (8) 36 (1964), no. 6, 816--818; see the source card. The statement and Lemme 1 are on p. 816, the proof on pp. 816--817.

Read depth. Claims checked: the statement, Lemme 1 and the case list of the proof were read clause by clause on the page images. The case analysis was recomputed here (see below); the theorem's proof was not otherwise reviewed, and nothing here is independently reviewed.

Proof pointer

Pp. 816--817. By Lemme 1 a pair nn, knkn needs a pseudoprime divisor nn of 2k−22^k-2. The note states that 2k−22^k-2 has no pseudoprime divisor for 1<k≤101<k\le10 and for k=13,14,18k=13,14,18. For k=12,16,20k=12,16,20 the number knkn would be divisible by 44, and no pseudoprime is. For each remaining k∈{11,15,17,19,21,22}k\in\{11,15,17,19,21,22\} the note lists the pseudoprime divisors of 2k−22^k-2 and observes that knkn is not a pseudoprime for any of them; the pair above settles k=23k=23.

For k=21k=21 the note's list gives only 11⋅3111\cdot31. A computation made for this page finds a second pseudoprime divisor of 221−22^{21}-2, namely 11⋅31⋅41=1398111\cdot31\cdot41=13981; the number 21⋅1398121\cdot13981 is not a pseudoprime, so the theorem is unaffected. The same computation confirms the other listed cases and that 6078760787 and 13981011398101 are pseudoprimes.

Bears on

  • Problem 649: the problem lists this note under the key [Ro64b], and the site's remarks cite that key for the statement that every prime p>13p>13 has a prime divisor q>pq>p of 2p−1−12^{p-1}-1. This theorem is not that statement and says nothing about the greatest prime factors of nn and n+1n+1.