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Statement

Notation (p. 206). A Sierpinski number is an odd integer $k$ such that $k\cdot2^n+1$ is composite for all $n\ge0$. A covering set for $k$ is a finite set of primes such that every $k\cdot2^n+1$, $n\ge0$, is divisible by at least one of them.

Theorem 1 (p. 206). Let the positive integer $t$ be any solution of the system of congruences

$$\begin{aligned} t&\equiv1\pmod 2,\\ t&\equiv1\text{ or }2\pmod 3,\\ t&\equiv0\pmod 5,\\ t&\equiv1,\ 4,\ 13\text{ or }16\pmod{17},\\ t&\equiv1,\ 16,\ 241\text{ or }256\pmod{257},\\ t&\equiv1,\ 256,\ 65281\text{ or }65536\pmod{65537},\\ t&\equiv1,\ 65536,\ 6634881\text{ or }6700416\pmod{6700417},\\ t&\equiv256,\ 318,\ 323\text{ or }385\pmod{641}. \end{aligned}$$

Then $k=t^4$ is a Sierpinski number.

The solutions $t$ form whole residue classes modulo the product of the eight moduli, so there are infinitely many such $t$ and infinitely many such $k$. The opening of the note (p. 206) says that it proves there are infinitely many Sierpinski numbers "of the new kind"; the print does not spell out this counting step.

Remark after the proof (p. 207). For these $k$ one has $k\cdot2^{4m+2}+1\equiv1\pmod 5$, so Sierpinski's set $\{3,5,17,257,641,65537,6700417\}$ is not a covering set for $k$. The note shows nothing more about covering sets for these $k$: it does not show that they have no finite covering set at all.

Question and suggestion (p. 207). The note then asks: "Are there other Sierpinski numbers analogous to Theorem 1?" Recalling the problem of the least $k_0$ with $k\cdot2^n+1$ always composite, with Selfridge's $78557$ and covering set $\{3,5,7,13,19,37,73\}$ the least known, it adds "Perhaps $k_0$ has no covering set." This is a suggestion, not a result.

Source. Anatoly S. Izotov, A note on Sierpiński numbers, Fibonacci Quart. 33 (1995), no. 3, 206-207: notation and Theorem 1 on p. 206, the proof on pp. 206-207, the remark, question and suggestion on p. 207. The edition read is identified on the source card.

Read depth. Claims checked: the statement, the remark and the question were read clause by clause on the printed pages, and the proof (pp. 206-207) was followed step by step. Nothing here is independently reviewed.

Proof sketch

Pp. 206-207. The congruences on $t$ give $k\equiv1$ modulo $2$, $3$, $17$, $257$, $65537$ and $6700417$, $k\equiv0\pmod 5$ and $k\equiv-1\pmod{641}$. Since $2$ has order $2$, $8$, $16$, $32$, $64$ and $64$ modulo $3$, $17$, $257$, $65537$, $6700417$ and $641$, with $2^{32}\equiv-1\pmod{641}$, the prime $3$ divides $k\cdot2^n+1$ for odd $n$, and $17$, $257$, $65537$, $6700417$ and $641$ divide it for $n\equiv4\pmod 8$, $n\equiv8\pmod{16}$, $n\equiv16\pmod{32}$, $n\equiv32\pmod{64}$ and $n\equiv0\pmod{64}$ respectively. That leaves $n=4m+2$, $m\ge0$, where Sophie Germain's identity gives

$$k\cdot2^{4m+2}+1=4(t\cdot2^m)^4+1 =\bigl(t^2 2^{2m+1}+t\,2^{m+1}+1\bigr)\bigl(t^2 2^{2m+1}-t\,2^{m+1}+1\bigr),$$

and both factors exceed $1$ because $t>1$. In the covered classes the value exceeds the dividing prime, since $t\equiv0\pmod5$ and the congruence modulo $6700417$ force $t\ge65536$; the printed proof leaves this step implicit.

Dependencies

None within the paper.

Bears on

• Problem 1113: the theorem gives infinitely many Sierpinski numbers for which composite values at $n\equiv2\pmod 4$ come from an algebraic factorization and not from a covering prime, and the remark shows that Sierpinski's seven-prime set does not cover them. It does not exhibit a Sierpinski number with no finite covering set, so it does not answer the problem.