A073503 Numbers n such that the number of solutions to x^4 == 1 (mod n) is twice the number of solutions of x^2 == 1 (mod n).

5, 10, 13, 15, 16, 17, 20, 25, 26, 29, 30, 32, 34, 35, 37, 39, 40, 41, 45, 48, 50, 51, 52, 53, 55, 58, 60, 61, 64, 68, 70, 73, 74, 75, 78, 82, 87, 89, 90, 91, 95, 96, 97, 100, 101, 102, 104, 105, 106, 109, 110, 111, 112, 113, 115, 116, 117, 119, 120, 122, 123, 125

Offset: 1

Author

Benoit Cloitre, Aug 19 2002

Keywords

• easy
• nonn

Comments

Conjectures: 2n > a(n) or 2n < a(n) for infinitely many values of n and abs(a(n)-2n) < sqrt(n) for n > 45. a(n)=2n for n = 318, 338, 350, 488, 490, 492, 494,...

Numbers divisible by 16 which have no prime factors = 1 mod 4, together with numbers not divisible by 16 which have exactly one prime factor = 1 mod 4. This refutes the conjectures. - Charles R Greathouse IV, Apr 16 2012

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A060594, A073103.

Programs

Mathematica

Select[Range[3, 125], Length[Reduce[x^4 - 1 == 0, x, Modulus -> #]] == 2*Length[Reduce[x^2 - 1 == 0, x, Modulus -> #]] &] (* Jayanta Basu, Jul 01 2013 *)

PARI

is(n)=my(v=factor(n)[,1]%4, s=sum(i=1,#v,v[i]==1), e=valuation(n, 2)); s==(e<4) \\ Charles R Greathouse IV, Apr 16 2012

Formula

a(n) seems to be asymptotic to 2n.