Cyclic quadratic residues

  • Thread starter ramsey2879
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I also wonder about an other interesting residue relation

Let P be a prime,

let [tex] a^{2^n}[/tex] be called a cyclic quadratic residue if there is integer m dependent on [tex]a[/tex] such that [tex] a^{2^{n + mp}} = a^{2^n}[/tex] for all integers [tex]p \mod P[/tex]
It seems that the sum of all such cylic residues is either 0 or 1 mod P
For instance for P = 17 the only cyclic residue is 1 but for P = 37
there are the cyclic sequences
33 16 34 9 7 12 33 ...
10 26 10 ...
1 ...
and the sum of all these numbers, not including repetitions is 4*37.
 
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  • #2
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Take the first 5 squares modulo 11: 1,4,9,5,3. Now square these numbers: Presto! We have them all back again 1,5,4,3,9. And the total is 22=11x2.
 
  • #3
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I also wonder about an other interesting residue relation

Let P be a prime,

let [tex] a^{2^n}[/tex] be called a cyclic quadratic residue if there is integer m dependent on [tex]a[/tex] such that [tex] a^{2^{n + mp}} = a^{2^n}[/tex] for all integers [tex]p \mod P[/tex]
It seems that the sum of all such cylic residues is either 0 or 1 mod P
For instance for P = 17 the only cyclic residue is 1 but for P = 37
there are the cyclic sequences
33 16 34 9 7 12 33 ...
10 26 10 ...
1 ...
and the sum of all these numbers, not including repetitions is 4*37.
Sorry I wasn't clear I mean let S(1) = k mod p, S(n) = S(n-1)^2 mod p. this sequence does not become cyclic until S(i) = S(i+j) and i > 1 so not all quadratic residues are necessarily cyclic quadratic residues. if p = 11 the cyclic sequences are 4,5,3,9,4,5,3,9 ... and 1,1,1,1,..
So all 5 residues are cyclic quadratic residues.
There are other related sequences such as S(n) = S(n-1)^2 - 2 which are also interesting because it is the sequence A(n) = F(2^n)/F(2^(n-1)) where F = the fibonacci sequences.
F(4) = 3 F(8) = 21, 21/3 = 7 = 3^2 -2 F(16)/F(8) = 987/21 = 47 = 7*7 - 2, etc which in mod 11 is 3,7,3,7,3,7,3,7,...., but in mod 17 is 3,7,13,14,7,13,14 .... where 7,13,and 14 sum to 34
 
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