Numbers n whose Euler Phi value Divides n + 1/Mistake/Third Edition

Source Work

 * $\mathbf B$: Divisibility
 * $\mathbf {B 37}$: Does $\map \phi n$ properly divide $n - 1$?
 * $\mathbf {B 37}$: Does $\map \phi n$ properly divide $n - 1$?

Mistake

 *  gives $8$ solutions to $\map \phi n \mid n + 1$, namely $n = 2$, $n = 2^{2^k} - 1$ for $1 \le k \le 5$, $n = n_1 = 3 \cdot 5 \cdot 17 \cdot 353 \cdot 929$ and $n = n_1 \cdot 83623937$. [Note that $353 = 11 \cdot 2^5 + 1, 929 = 29 \cdot 2^5 + 1, 83623937 = 11 \cdot 29 \cdot 2^{18} + 1$ and $\paren {353 - 2^8} \paren {929 - 2^8} = 2^{16} - 2^8 + 1$.] This exhausts the solutions with less than seven factors. notes that $n = n_1 \cdot 83623937 \cdot 699296672132097$ would be a solution were the largest factor a prime, put  notes that this is divisible by $73$.

Correction
While attributes that final observation to, it does in fact appear in 's original $1932$ article as a footnote:


 * ''In the same way if $6992962672132097$ is a prime
 * $3 \cdot 5 \cdot 17 \cdot 353 \cdot 929 \cdot 83623937 \cdot 6992962672132097 = 48901526933832864378258473353215$
 * is a solution of $(2)$.

Also note that $699296672132097$ is a misprint.

The correct number is $6992962672132097$.

Also see

 * Numbers $n$ whose Euler Phi value Divides $n + 1$