Note that Every Even perfect numbers (except 6 ) are

baih · 2019-08-14, 13:24

Note that Every Even perfect numbers (except 6 ) are :

2p−1(2p − 1) = 1 Mod (9*p) but not necessarily alternately.

p ======== perfect NUMBRE

3 ======== 28 =1 mod (27)
5 ======== 496 =1 mod (45)
7 ======== 8128 =1 mod (63)
13 ======== 33550336 =1 mod (117)
17 ======== 8589869056 =1 mod (153)
19 ======== 137438691328 =1 mod (171)
31 ======== 2305843008139952128 =1 mod (279)

is also work as primlity test :

p ======== 2p−1(2p − 1)

15 ======== 536854528 = 28 mod (135)
21 ======== 2199022206976 = 28 mod (189)
35 ======== 590295810341525782528 = 118 mod (315)

Dr Sardonicus · 2019-08-14, 13:46

If p is prime, then 2^p-1 == 1 (mod p) and 2^p == 2 (mod p) so 2^p-1(2^p - 1) == 1*1 == 1 (mod p).

If p > 3, then p == 1 or 5 (mod 6).

If p == 1 (mod 6) then 2^p-1 == 1 (mod 9) and 2^p - 1 == 1 (mod 9), so 2^p-1(2^p - 1) == 1*1 == 1 (mod 9).

If p == 5 (mod 6) then 2^p-1 == 7 (mod 9) and 2^p - 1 == 4 (mod 9), so 2^p-1(2^p - 1) == 7*4 == 1 (mod 9).

Another triumph for elementary number theory!

baih · 2019-08-14, 13:54

lol

2019-08-14, 13:24	#1
baih Jun 2019 100010₂ Posts	Note that Every Even perfect numbers (except 6 ) are Note that Every Even perfect numbers (except 6 ) are : 2p−1(2p − 1) = 1 Mod (9p) but not necessarily alternately. p ======== perfect NUMBRE 3 ======== 28 =1 mod (27) 5 ======== 496 =1 mod (45) 7 ======== 8128 =1 mod (63) 13 ======== 33550336 =1 mod (117) 17 ======== 8589869056 =1 mod (153) 19 ======== 137438691328 =1 mod (171) 31 ======== 2305843008139952128 =1 mod (279) is also work as primlity test : p ======== 2p−1(2p − 1) 15 ======== 536854528 = 28 mod (135) 21 ======== 2199022206976 = 28 mod (189) 35 ======== 590295810341525782528 = 118 mod (315) Last fiddled with by baih on 2019-08-14 at 13:28*

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2019-08-14, 13:46	#2
Dr Sardonicus Feb 2017 Nowhere 13×17×29 Posts	If p is prime, then 2^p-1 == 1 (mod p) and 2^p == 2 (mod p) so 2^p-1(2^p - 1) == 11 == 1 (mod p). If p > 3, then p == 1 or 5 (mod 6). If p == 1 (mod 6) then 2^p-1 == 1 (mod 9) and 2^p - 1 == 1 (mod 9), so 2^p-1(2^p - 1) == 11 == 1 (mod 9). If p == 5 (mod 6) then 2^p-1 == 7 (mod 9) and 2^p - 1 == 4 (mod 9), so 2^p-1(2^p - 1) == 7*4 == 1 (mod 9). Another triumph for elementary number theory!