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While researching for a post on tetranacci pseudoprimes I came across a list of Carmichael numbers,

$$C_n = 561,\, 1105,\, 1729,\, 2465,\, 2821,\dots$$

Of course, Ramanujan's taxicab number $1729 = 12^3+1$ stood out. So I looked at other near-cubes $C(m) = m^3+1$ and found that for $n<10000$, there were six, namely,

$$m = 12,\, 36,\, 138,\, 270,\, 4800,\, 7560,\dots?$$

$C(m) = m^3+1 = (m+1)(m^2-m+1)$ have the factorizations,

$$C(12) = 7 \cdot \color{brown}{13} \cdot 19$$

$$C(36) = 13 \cdot \color{brown}{37} \cdot 97$$

$$C(138) = 7 \cdot 37 \cdot 73 \cdot \color{brown}{139}$$

$$C(270) = 13 \cdot 37 \cdot 151 \cdot \color{brown}{271}$$

$$C(4800) = 7 \cdot 19 \cdot 31 \cdot 37 \cdot 151 \cdot \color{brown}{4801}$$

$$C(7560) = 271 \cdot 433 \cdot 487 \cdot \color{brown}{7561}$$

Note that $m+1$ is prime.

Questions:

  1. Since there are an infinite number of Carmichael numbers, what others are of form $m^3+1$?
  2. Is it true that if it is of form $m^3+1$, then $m+1$ is prime?

P.S. Richard Pinch's database with $C_n<10^{21}$ seem to have broken links.

$\color{brown}{Update}$: Courtesy of R. Pinch's answer, the complete list with $C_n<10^{21}$ include three more,

$$C(12840)$$

$$C(14700)$$

$$C(678480)$$

and only $14700+1$ is composite.

Community
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Tito Piezas III
  • 54,565

1 Answers1

6

The first few Carmichael numbers of the form $m^3+1$ are

 1729
 46657
 2628073
 19683001
 110592000001
 432081216001
 2116874304001
 3176523000001
 312328165704192001

The eighth in the list, which is $C_{12616}$, has a composite value of $m+1$.

BTW, my tables are currently at http://s369624816.websitehome.co.uk/

Richard Pinch
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