

A206549


Nontrivial solutions of x^2 == 1 (Modd p), p an odd prime, for odd restricted residue classes Modd p.


6



3, 5, 13, 17, 31, 9, 23, 11, 27, 55, 75, 91, 33, 15, 37, 105, 129, 93, 19, 81, 183, 107, 89, 177, 241, 187, 217, 53, 155, 25, 203, 189, 213, 311, 269, 115, 63, 381, 143, 29, 179, 67, 109, 413, 301, 235, 489, 439, 483, 553
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OFFSET

1,1


COMMENTS

For multiplication Modd n (not to be confused with multiplication mod n) see a comment on A203571.
The trivial solution of x^2 == 1 (Modd n) is x = 1 (Modd n). Note that x = 1 (Modd n) == +1 (Modd n). In the ordinary mod n case the trivial solution is 1 (mod 2) for n=2 (1 ==+1(mod 2)) and if n>2 the two trivial solutions are 1 (mod n) and the noncongruent 1 (mod n) == n1 (mod n).
Here multiplication on the reduced residue system Modd p, p an odd prime, with only odd numbers is considered (which is possible, contrary to mod p). In order to have inverses one has to exclude all reduced residue classes Modd p with even numbers. The (p1)/2 residue classes are then [1],[3],,..., [p2]. For m=1,3,...,p2, the class [m] is the union of the ordinary reduced residue classes mod 2p: [m] and [m]=[2pm]. Besides the trivial solution x=+1 (Modd p) (note that 1 == +1 (Modd p)) there is a further nontrivial solution if and only if (p1)/2 is even, i.e. p=p(n)=A002144(n) (primes of the form 4*k+1), n>=1. The present sequence entry a(n) gives the smallest positive representative for this nontrivial solution
Modd A002144(n).
This result uses the fact that every finite group of prime order p is the cyclic group Z_p (Corollary to Lagrange's or also Cauchy's theorem on finite groups or see A000688 for abelian groups). Here for the multiplicative group Modd p (on the odd residue classes) which has order (p1)/2, for p an odd prime. This turns out to be the Galois group for the minimal polynomial C(p,x), whose coefficients are found in A187360. The sequence {a(n)} arises if one asks for the smallest positive members of the reduced residue system Modd p, namely [1], [3],...[p2], which are their own inverses besides the trivial element 1 (Modd p).
The row a(n) has in the table for the multiplicative group Modd A002144(n) a 1 on the diagonal, if the table is written for the odd representatives 1,3,...,p2. The only other diagonal entry 1 appears for the element 1. Note that in the ordinary mod p case, p an odd prime, one always has for the multiplicative group mod p (which has p1 residue classes) in the multiplication table the diagonal entry 1 only for the representatives 1 and p1, but p1 == 1 (mod p) is a trivial solution of x^2 == 1 (mod p).


LINKS

Table of n, a(n) for n=1..50.


FORMULA

a(n)^2 == 1 Modd(A002144(n)), n>=1, a(n) the smallest positive solution not 1. For Modd p, p an odd prime, see the comment section and the examples.


EXAMPLE

a(6)=9 because the corresponding prime is A002144(6)=41, 9^2 = 81, and 81 (Modd 41) is per definition 81 (mod 41) = +1 (the definition uses the parity of floor(81/41) = 1 being odd, hence the  sign), thus 9^2 == 1 Modd(41), and 9 is not congruent 1 (Modd 41) (or 1 (Modd 41)), hence a nontrivial solution.
A002144(n): 5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, ...
a(n): 3, 5, 13, 17, 31, 9, 23, 11, 27, 55, 75, ...
3^2 = 9, 9 (Mod 5) := 9 (mod 10) = 1 (the smallest positive representative of the class 1 (Modd 5) = {+/1,+/9,+/10,+/19,..).
5^2 = 25, 25 (Modd 13) := 13 (mod 26) = 1.
13^2 = 169, 169 (Modd 17) := 169 (mod 34) = 1.
17^2 = 289, 289 (Modd 29) := 289 (mod 2*29) = 1.
...
E.g., for the prime 7, not in A002144, there are no self inverse elements in the multiplicative group Modd 7 (on the odd numbers) except the trivial 1. The inverse of 3 is 5 (Modd 7), and vice versa.


CROSSREFS

Sequence in context: A073654 A060192 A065311 * A040158 A147490 A266234
Adjacent sequences: A206546 A206547 A206548 * A206550 A206551 A206552


KEYWORD

nonn


AUTHOR

Wolfdieter Lang, Feb 13 2012


STATUS

approved



