Wikipedia:Reference desk/Archives/Mathematics/2020 February 13

= February 13 =

Even perfect numbers in duodecimal
Is it true that every even perfect number other than 6 and 24 in duodecimal ends with the digits "54"? Remember that while the duodecimal number 24 may look like the number twenty-four, it really means twenty-eight, which, of course, we know is a perfect number.

As an example, consider the perfect number 496 (in decimal). In duodecimal, it becomes 354 (not to be confused with the decimal number three hundred and fifty-four). Notice that the last two digits here are "54". GeoffreyT2000 (talk) 06:01, 13 February 2020 (UTC)


 * Yes, it is true. Here is a simpler statement, not involving primality, that implies the one above about perfect numbers:
 * For all odd p, p ≥ 5, 22p−1 − 2p−1 ≡ 26 (mod 144).
 * This statement can easily be checked by checking it separately (using modular arithmetic) for each of the three congruence classes of odd p modulo 6. Note that 26·26 ≡ 26 (mod 144). --Lambiam 08:39, 13 February 2020 (UTC)


 * The necessary arithmetic calculations can be somewhat simplified:
 * 22p−1 − 2p−1 ≡ 26 (mod 144)
 * ⇔ 22p+5 − 2p+5 ≡ 26 (mod 144)
 * ⇔ 24·(22p+1 − 2p+1) ≡ 24·4 (mod 24·9)
 * ⇔ 22p+1 − 2p+1 ≡ 4 (mod 9)
 * ⇔ 4·(22p−1 − 2p−1) ≡ 4 (mod 9)
 * ⇐ 22p−1 − 2p−1 ≡ 1 (mod 9).
 * So the original statement follows by modular arithmetic from the fact that sufficiently large (≥ 28) even perfects end on a digit 1 when expressed nonally. I see no obvious way of avoiding case analysis. --Lambiam 13:58, 13 February 2020 (UTC)


 * Without case analysis. Let v = 2p−1. Since p is odd, we can write it in the form 2q + 1, so then v = 4q ≡ 1 (mod 3). Hence, v can be written in the form 3u + 1. Now, by elementary algebra,
 * 22p−1 − 2p−1
 * = (2p−1)(2·2p−1 − 1)
 * = v(2v − 1)
 * = (3u + 1)(6u + 1)
 * = 18u2 + 9u + 1
 * ≡ 1 (mod 9).
 * --Lambiam 19:20, 13 February 2020 (UTC)