The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A372953 (Bold, blue-underlined text is an addition; faded, red-underlined text is a ~~deletion~~.)
Showing entries 1-10 | older changes

A372953

Orders of finite fields where -1 is a square.
(history; published version)

#16 by Michael De Vlieger at Sat Jun 01 06:17:28 EDT 2024

STATUS

reviewed

approved

#15 by Joerg Arndt at Sat Jun 01 05:47:57 EDT 2024

STATUS

proposed

reviewed

#14 by Mike Speciner at Sun May 19 09:50:32 EDT 2024

STATUS

editing

proposed

#13 by Mike Speciner at Sun May 19 09:44:55 EDT 2024

COMMENTS

The sequence comprises the positive powers of 2, the following: 2, positive powers of primes congruent to 1 mod 4, and squares the positive even powers of primes congruent to 3 mod 4.

STATUS

proposed

editing

Discussion

Sun May 19

09:49

Mike Speciner: Revised comment again. Hopefully it's unambiguous now.

#12 by Mike Speciner at Sat May 18 22:46:47 EDT 2024

STATUS

editing

proposed

Discussion

Sun May 19

01:50

Joerg Arndt: "squares of primes congruent to 3 mod 4" is misleading as I said; rather say  "powers with even exponents of primes congruent to 3 mod 4" ?

#11 by Mike Speciner at Sat May 18 22:45:00 EDT 2024

COMMENTS

The multiplication group of GF(p^n) is cyclic of order o = p^n-1. For p=2, 1=-1, so 1 is a square root of -1. Otherwise, -1 has order 2 and so any square root of -1 has order 4. So, for there to be a square root of -1, o mod 4 must be 0, i.e. p^n mod 4 = 1. Then if g is a generator of the group, g^(o/4) is a square root of -1. p^n mod 4 = 1 if and only if p mod 4 = 1 or p mod 4 = 3 and n is even.

CROSSREFS

Symmetric difference of A000079 (power of 2) and A085759 (prime powers congruent to 1 mod 4).

STATUS

proposed

editing

#10 by Michel Marcus at Sat May 18 13:43:30 EDT 2024

STATUS

editing

proposed

Discussion

Sat May 18

18:07

Robert P. P. McKone: This sequence appears to contain all of A002313.

18:17

Mike Speciner: Yes, it contains all of A002313, which are the primes congruent to 1 or 2 mod 4. And it contains all higher powers of said primes as well. Of course 2 is the only prime congruent to 2 mod 4.

#9 by Michel Marcus at Sat May 18 13:43:01 EDT 2024

DATA

2, 4, 5, 8, 9, 13, 16, 17, 25, 29, 32, 37, 41, 49, 53, 61, 64, 73, 81, 89, 97, 101, 109, 113, 121, 125, 128, 137, 149, 157, 169, 173, 181, 193, 197, 229, 233, 241, 256, 257, 269, 277, 281, 289, 293, 313, 317, 337, 349, 353, 361, 373, 389, 397, 401, 409, 421, 433, 449, 457, 461

STATUS

proposed

editing

Discussion

Sat May 18

13:43

Michel Marcus: 3 lines 1/2 is usually ok

#8 by Mike Speciner at Sat May 18 06:51:27 EDT 2024

STATUS

editing

proposed

Discussion

Sat May 18

07:28

Robert P. P. McKone: I think a few examples of these finite fields would need to be added to have this sequence confirmed and make sense?

08:26

Mike Speciner: The multiplicative group of GF(p^n) has order p^n-1. If p=1 mod 4, that order is a multiple of 4, while if p=3 mod 4 that order is a multiple of 4 only when n is even. If p = 2, 1=-1, so -1 is always a square; otherwise -1 has order 2 so its square root must have order 4 which must divide the multiplicative order of the field. So when -1 is not a square, i.e. when p=3 mod 4 and n is odd, it is in the splitting field of x^2+1, i.e. in GF(p^(2n)). Perhaps see A008784 to see the GF(p) case. From that it's easy to see that either GF(p) or GF(p^2) has a square root of -1, and so any field extension also has a square root of -1.

08:34

Mike Speciner: Perhaps I should have mentioned that the multiplicative group of a finite field is cyclic, which implies that 1 has a kth root for any k that divides the order of the group.

#7 by Mike Speciner at Sat May 18 06:49:50 EDT 2024

PROG

(Python)

(Python)from itertools import count