cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 11 results. Next

A002145 Primes of the form 4*k + 3.

Original entry on oeis.org

3, 7, 11, 19, 23, 31, 43, 47, 59, 67, 71, 79, 83, 103, 107, 127, 131, 139, 151, 163, 167, 179, 191, 199, 211, 223, 227, 239, 251, 263, 271, 283, 307, 311, 331, 347, 359, 367, 379, 383, 419, 431, 439, 443, 463, 467, 479, 487, 491, 499, 503, 523, 547, 563, 571
Offset: 1

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Author

Keywords

Comments

Or, odd primes p such that -1 is not a square mod p, i.e., the Legendre symbol (-1/p) = -1. [LeVeque I, p. 66]. - N. J. A. Sloane, Jun 28 2008
Primes which are not the sum of two squares, see the comment in A022544. - Artur Jasinski, Nov 15 2006
Natural primes which are also Gaussian primes. (It is a common error to refer to this sequence as "the Gaussian primes".)
Inert rational primes in the field Q(sqrt(-1)). - N. J. A. Sloane, Dec 25 2017
Numbers n such that the product of coefficients of (2n)-th cyclotomic polynomial equals -1. - Benoit Cloitre, Oct 22 2002
For p and q both belonging to the sequence, exactly one of the congruences x^2 = p (mod q), x^2 = q (mod p) is solvable, according to Gauss reciprocity law. - Lekraj Beedassy, Jul 17 2003
Also primes p that divide L((p-1)/2) or L((p+1)/2), where L(n) = A000032(n), the Lucas numbers. Union of A122869 and A122870. - Alexander Adamchuk, Sep 16 2006
Also odd primes p that divide ((p-1)!! + 1) or ((p-2)!! + 1). - Alexander Adamchuk, Nov 30 2006
Also odd primes p that divide ((p-1)!! - 1) or ((p-2)!! - 1). - Alexander Adamchuk, Apr 18 2007
This sequence is a proper subset of the set of the absolute values of negative fundamental discriminants (A003657). - Paul Muljadi, Mar 29 2008
Bernard Frénicle de Bessy discovered that such primes cannot be the hypotenuse of a Pythagorean triangle in opposition to primes of the form 4*n+1 (see A002144). - after Paul Curtz, Sep 10 2008
A079261(a(n)) = 1; complement of A145395. - Reinhard Zumkeller, Oct 12 2008
Subsequence of A007970. - Reinhard Zumkeller, Jun 18 2011
A151763(a(n)) = -1.
Primes p such that p XOR 2 = p - 2. Brad Clardy, Oct 25 2011 (Misleading in the sense that this is a formula for the super-sequence A004767. - R. J. Mathar, Jul 28 2014)
It appears that each term of A004767 is the mean of two terms of this subsequence of primes therein; cf. A245203. - M. F. Hasler, Jul 13 2014
Numbers n > 2 such that ((n-2)!!)^2 == 1 (mod n). - Thomas Ordowski, Jul 24 2016
Odd numbers n > 1 such that ((n-1)!!)^2 == 1 (mod n). - Thomas Ordowski, Jul 25 2016
Primes p such that (p-2)!! == (p-3)!! (mod p). - Thomas Ordowski, Jul 28 2016
See Granville and Martin for a discussion of the relative numbers of primes of the form 4k+1 and 4k+3. - Editors, May 01 2017
Sometimes referred to as Blum primes for their connection to A016105 and the Blum Blum Shub generator. - Charles R Greathouse IV, Jun 14 2018
Conjecture: a(n) for n > 4 can be written as a sum of 3 primes of the form 4k+1, which would imply that primes of the form 4k+3 >= 23 can be decomposed into a sum of 6 nonzero squares. - Thomas Scheuerle, Feb 09 2023

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 870.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 146-147.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 5th ed., Oxford Univ. Press, 1979, p. 219, th. 252.
  • W. J. LeVeque, Topics in Number Theory. Addison-Wesley, Reading, MA, 2 vols., 1956, Vol. 1, p. 66.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 90.

Crossrefs

Apart from initial term, same as A045326.
Cf. A016105.
Cf. A004614 (multiplicative closure).

Programs

  • Haskell
    a002145 n = a002145_list !! (n-1)
    a002145_list = filter ((== 1) . a010051) [3, 7 ..]
    -- Reinhard Zumkeller, Aug 02 2015, Sep 23 2011
    
  • Magma
    [4*n+3 : n in [0..142] | IsPrime(4*n+3)]; // Arkadiusz Wesolowski, Nov 15 2013
    
  • Maple
    A002145 := proc(n)
        option remember;
        if n = 1 then
            3;
        else
            a := nextprime(procname(n-1)) ;
            while a mod 4 <>  3 do
                a := nextprime(a) ;
            end do;
            return a;
        end if;
    end proc:
    seq(A002145(n),n=1..20) ; # R. J. Mathar, Dec 08 2011
  • Mathematica
    Select[4Range[150] - 1, PrimeQ] (* Alonso del Arte, Dec 19 2013 *)
    Select[ Prime@ Range[2, 110], Length@ PowersRepresentations[#^2, 2, 2] == 1 &] (* or *)
    Select[ Prime@ Range[2, 110], JacobiSymbol[-1, #] == -1 &] (* Robert G. Wilson v, May 11 2014 *)
  • PARI
    forprime(p=2,1e3,if(p%4==3,print1(p", "))) \\ Charles R Greathouse IV, Jun 10 2011
    
  • Sage
    def A002145_list(n): return [p for p in prime_range(1, n + 1) if p % 4 == 3]  # Peter Luschny, Jul 29 2014

Formula

Remove from A000040 terms that are in A002313.
Intersection of A000040 and A004767. - Alonso del Arte, Apr 22 2014
From Vaclav Kotesovec, Apr 30 2020: (Start)
Product_{k>=1} (1 - 1/a(k)^2) = A243379.
Product_{k>=1} (1 + 1/a(k)^2) = A243381.
Product_{k>=1} (1 - 1/a(k)^3) = A334427.
Product_{k>=1} (1 + 1/a(k)^3) = A334426.
Product_{k>=1} (1 - 1/a(k)^4) = A334448.
Product_{k>=1} (1 + 1/a(k)^4) = A334447.
Product_{k>=1} (1 - 1/a(k)^5) = A334452.
Product_{k>=1} (1 + 1/a(k)^5) = A334451. (End)
From Vaclav Kotesovec, May 05 2020: (Start)
Product_{k>=1} (1 + 1/a(k)) / (1 + 1/A002144(k)) = Pi/(4*A064533^2) = 1.3447728438248695625516649942427635670667319092323632111110962...
Product_{k>=1} (1 - 1/a(k)) / (1 - 1/A002144(k)) = Pi/(8*A064533^2) = 0.6723864219124347812758324971213817835333659546161816055555481... (End)
Sum_{k >= 1} 1/a(k)^s = (1/2) * Sum_{n >= 1 odd numbers} moebius(n) * log(2 * (2^(n*s) - 1) * (n*s - 1)! * zeta(n*s) / (Pi^(n*s) * abs(EulerE(n*s - 1))))/n, s >= 3 odd number. - Dimitris Valianatos, May 20 2020

Extensions

More terms from James Sellers, Apr 21 2000

A385192 Primes p such that multiplicative order of 5 modulo p is odd.

Original entry on oeis.org

2, 11, 19, 31, 59, 71, 79, 101, 109, 131, 139, 149, 151, 179, 181, 191, 199, 211, 239, 251, 269, 271, 311, 331, 359, 379, 389, 401, 409, 419, 431, 439, 461, 479, 491, 499, 541, 569, 571, 599, 619, 631, 659, 691, 719, 739, 751, 811, 829, 839, 859, 911, 919, 941, 971, 991
Offset: 1

Views

Author

Jianing Song, Jun 20 2025

Keywords

Comments

The multiplicative order of 5 modulo a(n) is A385193(n).
Contained in primes congruent to 1 or 4 modulo 5 (primes p such that 5 is a quadratic residue modulo p, A045468), and contains primes congruent to 11 or 19 modulo 20 (A122869).
Conjecture: this sequence has density 1/3 among the primes.

Examples

			101 is a term since 5^25 == 1 (mod 101).
		

Crossrefs

Subsequence of A040105, which (without the terms 2 and 5) is itself a subsequence of A045468.
Contains A122869 as a proper subsequence.
Cf. A385193 (the actual multiplicative orders).
Cf. other bases: A014663 (base 2), A385220 (base 3), A385221 (base 4), this sequence (base 5), A163183 (base -2), A385223 (base -3), A385224 (base -4), A385225 (base -5).

Programs

  • Mathematica
    Select[Prime[Range[200]], OddQ[MultiplicativeOrder[5, #]] &] (* Paolo Xausa, Jun 28 2025 *)
  • PARI
    isA385192(p) = isprime(p) && (p!=5) && znorder(Mod(5,p))%2

A122870 Primes congruent to 3 or 7 mod 20.

Original entry on oeis.org

3, 7, 23, 43, 47, 67, 83, 103, 107, 127, 163, 167, 223, 227, 263, 283, 307, 347, 367, 383, 443, 463, 467, 487, 503, 523, 547, 563, 587, 607, 643, 647, 683, 727, 743, 787, 823, 827, 863, 883, 887, 907, 947, 967, 983, 1063, 1087, 1103, 1123, 1163, 1187, 1223
Offset: 1

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Author

Alexander Adamchuk, Sep 16 2006

Keywords

Comments

The old name was "Primes p that divide Lucas((p+1)/2) = A000032((p+1)/2)".
Note that F(p+1) = F((p+1)/2)*Lucas((p+1)/2), where F = A000045. Since gcd(F(n),Lucas(n)) = 1 or 2 (because Lucas(n)^2 - 5*F(n)^2 = 4*(-1)^n), this sequence (under the old definition above) lists primes p such that p divides F(p+1) but does not divides F((p+1)/2). By Propositions 1.1 and 1.2 (the k = 3 case) of my link below, this is primes p == 3, 7 (mod 20). - Jianing Song, Jun 20 2025

References

  • David A. Cox, "Primes of the Form x^2 + n y^2", Wiley, 1989; see p. 33.

Crossrefs

Subseqeunce of A002145, A003631, A049098, A053027. Essentially the same as A106865.

Programs

  • Magma
    [p: p in PrimesUpTo(1500) | p mod 20 in [3, 7]]; // Vincenzo Librandi, Jan 06 2013
  • Mathematica
    Select[Prime[Range[1000]],IntegerQ[(Fibonacci[(#1+1)/2-1]+Fibonacci[(#1+1)/2+1])/#1]&]
    Select[Prime[Range[300]], MemberQ[{3, 7}, Mod[#, 20]]&] (* Vincenzo Librandi, Jan 06 2013 *)

Extensions

I merged A216816 into this entry at the suggestion of Jianing Song, Jun 20 2025. - N. J. A. Sloane, Jun 22 2025

A385167 Primes p == 3 (mod 4) such that (p+1) * ord(5,p) / ord(2+-i,p) is even. Here ord(a,m) is the multiplicative order of a modulo m.

Original entry on oeis.org

11, 79, 131, 199, 211, 239, 251, 331, 359, 439, 479, 491, 571, 599, 691, 719, 811, 839, 919, 971, 1039, 1051, 1091, 1171, 1279, 1291, 1319, 1399, 1439, 1451, 1531, 1559, 1571, 1759, 1811, 1879, 1931, 1999, 2011, 2039, 2131, 2239, 2251, 2371, 2399, 2411, 2531, 2719, 2731, 2851, 2879, 2971, 2999
Offset: 1

Views

Author

Jianing Song, Jun 20 2025

Keywords

Comments

Of course if a and m are integers, it doesn't matter if the base ring is Z or Z[i] for ord(a,m).
List of p = A002145(k) such that A385166(k) is even.
Since in this case d(p) divides (p^2-1)/2, 5 must be a quadratic residue modulo p (see A385165).

Examples

			359 is a term since the multiplicative order of 2+-i modulo 359 is 6444, and (360*ord(5,359))/6444 = 10 is even.
		

Crossrefs

Cf. A002145, A385165 (list of ord(2+-i,p)), A385166 (list of (p+1) * ord(5,p) / ord(2+-i,p)).
Subsequence of the intersection of A122869 and A385168. Contains A385180 as a subsequence.

Programs

  • PARI
    quot(p) = my(z = znorder(Mod(5,p)), d = divisors((p+1)*z)); for(i=1, #d, if(Mod([2,-1;1,2],p)^d[i] == 1, return((p+1)*z/d[i]))) \\ for a prime p == 3 (mod 4), returns (p+1) * ord(5,p) / ord(2+-i, p)
    isA385167(p) = isprime(p) && p%4==3 && quot(p)%2==0

A385180 Primes p == 3 (mod 4) such that (p+1) * ord(5,p) / ord(2+-i,p) is divisible by 4. Here ord(a,m) is the multiplicative order of a modulo m.

Original entry on oeis.org

331, 571, 599, 691, 839, 919, 971, 1039, 1051, 1171, 1279, 1291, 1319, 1399, 1439, 1451, 1571, 1759, 1879, 2131, 2411, 2879, 2971, 3079, 3251, 3331, 3491, 3571, 3691, 3851, 4051, 4079, 4091, 4211, 4519, 4639, 4651, 4679, 4691, 4759, 4919, 4931, 5051, 5119, 5171, 5279, 5479, 5519, 5531
Offset: 1

Views

Author

Jianing Song, Jun 20 2025

Keywords

Comments

Of course if a and m are integers, it doesn't matter if the base ring is Z or Z[i] for ord(a,m).
List of p = A002145(k) such that A385166(k) is divisible by 4.
Since in this case d(p) divides (p^2-1)/2, 5 must be a quadratic residue modulo p (see A385165).
By definition, a term that is in neither A385169 nor A385179 must be congruent to 31 or 79 modulo 80. The smallest such term is p = 1759 (ord(2+-i,p) = ((p+1)/4) * ord(5,p) = 128920); even if 1039 == 79 (mod 80), we have ord(2+-i,p) = ((p+1)/8) * ord(5,p) = 22490 == 2 (mod 4), which means that 1039 is in A385179.

Examples

			571 is a term since the multiplicative order of 2+-i modulo 571 is 40755, and (572*ord(5,571))/40755 = 4 is divisible by 4.
		

Crossrefs

Cf. A002145, A385165 (list of ord(2+-i,p)), A385166 (list of (p+1) * ord(5,p) / ord(2+-i,p)).
Subsequence of A385167, which is itself a subsequence of intersection of A122869 and A385168.

Programs

  • PARI
    quot(p) = my(z = znorder(Mod(5,p)), d = divisors((p+1)*z)); for(i=1, #d, if(Mod([2,-1;1,2],p)^d[i] == 1, return((p+1)*z/d[i]))) \\ for a prime p == 3 (mod 4), returns (p+1) * ord(5,p) / ord(2+-i, p)
    isA385180(p) = isprime(p) && p%4==3 && quot(p)%4==0

A385169 Primes p == 3 (mod 4) such that the multiplicative order of 2+-i modulo p in Gaussian integers (A385165) is odd.

Original entry on oeis.org

331, 571, 599, 691, 839, 971, 1051, 1171, 1291, 1451, 1571, 1879, 2131, 2411, 2971, 3251, 3331, 3491, 3571, 3691, 3851, 4051, 4091, 4211, 4651, 4679, 4691, 4919, 4931, 5051, 5171, 5479, 5531, 5651, 5839, 5851, 5879, 6011, 6599, 6679, 6691, 7079, 7211, 7331, 7691, 8011, 8039, 8171, 8731, 8839, 9011, 9371, 9811
Offset: 1

Views

Author

Jianing Song, Jun 20 2025

Keywords

Comments

Primes p == 3 (mod 4) are precisely the rational primes in the ring of Gaussian integers.
Let ord(a,m) be the multiplicative order of a modulo m. (Of course if a and m are integers, it doesn't matter if the base ring is Z or Z[i]). For a prime p == 3 (mod 4), we have that ord(2+-i,p) is divisible by ord(5,p), and that ord(2+-i,p) divides (p+1) * ord(5,p). What's more, ord(2+-i,p) divides (p^2-1)/2 if and only if 5 is a quadratic residue of integers modulo p. (See A385165).
As a result, if ord(2+-i,p) is not divisible by 8, then ord(5,p) is odd:
- Of course this is true if ord(2+-i,p) is odd.
- If ord(2+-i,p) == 2 (mod 4) and ord(5,p) is even, then ord(2+-i,p)/ord(5,p) is odd, and so ord(2+-i,p) divides ((p+1)/4) * ord(5,p), then ord(5,p) is odd. This implies that ord(2+-i,p) is odd, a contradiction.
- If ord(2+-i,p) == 4 (mod 8) and ord(5,p) is even (we have ord(5,p) == 2 (mod 4) since p == 3 (mod 4)), then ord(2+-i,p)/ord(5,p) == 2 (mod 4), and so ord(2+-i,p) divides ((p+1)/2) * ord(5,p), then ord(5,p) is odd. This implies that ord(2+-i,p) == 2 (mod 4), a contradiction.
From the above paragraph, this sequence is also primes p == 3 (mod 4) such that ord(2+-i,p)/ord(5,p) is odd.

Examples

			8731 is a term since (2+-i)^635253 == 1 (mod 8731), and 635253 is odd.
8839 is a term since (2+-i)^57447 == 1 (mod 8839), and 57447 is odd.
9011 is a term since (2+-i)^2029953 == 1 (mod 9011), and 2029953 is odd.
		

Crossrefs

Cf. A385165, A385179, A385192, A385217 (the actual multiplicative orders).
A385188 < this sequence < A385180 < A385167 < intersection of A122869 and A385168, where Ax < Ay means that Ax is a subsequence of Ay.

Programs

  • PARI
    ord(p) = my(d = divisors((p+1)*znorder(Mod(5, p)))); for(i=1, #d, if(Mod([2, -1; 1, 2], p)^d[i] == 1, return(d[i]))) \\ for a prime p == 3 (mod 4), returns ord(2+-i, p)
    isA385169(p) = isprime(p) && p%4==3 && ord(p)%2

A385179 Primes p == 3 (mod 4) such that the multiplicative order of 2+-i modulo p in Gaussian integers (A385165) is congruent to 2 modulo 4.

Original entry on oeis.org

11, 131, 211, 251, 491, 811, 919, 1039, 1091, 1319, 1399, 1531, 1811, 1931, 2011, 2251, 2371, 2531, 2731, 2851, 3011, 3079, 3371, 3931, 4079, 4451, 4519, 4759, 5011, 5639, 6091, 6131, 6211, 6359, 6451, 6491, 6571, 6971, 7411, 7451, 7559, 7639, 8291, 8719, 8971, 9091, 9491, 9719, 9839, 9851, 9931
Offset: 1

Views

Author

Jianing Song, Jun 20 2025

Keywords

Comments

Primes p == 3 (mod 4) are precisely the rational primes in the ring of Gaussian integers.
Let ord(a,m) be the multiplicative order of a modulo m. (Of course if a and m are integers, it doesn't matter if the base ring is Z or Z[i]). For a prime p == 3 (mod 4), we have that ord(2+-i,p) is divisible by ord(5,p), and that ord(2+-i,p) divides (p+1) * ord(5,p). What's more, ord(2+-i,p) divides (p^2-1)/2 if and only if 5 is a quadratic residue of integers modulo p. (See A385165).
As a result, if ord(2+-i,p) is not divisible by 8, then ord(5,p) is odd:
- Of course this is true if ord(2+-i,p) is odd.
- If ord(2+-i,p) == 2 (mod 4) and ord(5,p) is even, then ord(2+-i,p)/ord(5,p) is odd, and so ord(2+-i,p) divides ((p+1)/4) * ord(5,p), then ord(5,p) is odd. This implies that ord(2+-i,p) is odd, a contradiction.
- If ord(2+-i,p) == 4 (mod 8) and ord(5,p) is even (we have ord(5,p) == 2 (mod 4) since p == 3 (mod 4)), then ord(2+-i,p)/ord(5,p) == 2 (mod 4), and so ord(2+-i,p) divides ((p+1)/2) * ord(5,p), then ord(5,p) is odd. This implies that ord(2+-i,p) == 2 (mod 4), a contradiction.
From the above paragraph, this sequence is also primes p == 3 (mod 4) such that ord(2+-i,p)/ord(5,p) == 2 (mod 4).

Examples

			919 is a term since (2+-i)^21114 == 1 (mod 919), (2+-i)^(21114/2) !== 1 (mod 919), and we have 21114 == 2 (mod 4).
		

Crossrefs

Cf. A385165, A385169, A385188, A385218 (the actual multiplicative orders).
Subsequence of A385167, which itself lies in the intersection of A122869 and A385168.

Programs

  • PARI
    ord(p) = my(d = divisors((p+1)*znorder(Mod(5, p)))); for(i=1, #d, if(Mod([2, -1; 1, 2], p)^d[i] == 1, return(d[i]))) \\ for a prime p == 3 (mod 4), returns ord(2+-i, p)
    isA385179(p) = isprime(p) && p%4==3 && ord(p)%4==2

A385188 Primes p == 3 (mod 4) such that the multiplicative order of 2+-i modulo p in Gaussian integers (A385165) is not divisible by 2 or 3.

Original entry on oeis.org

599, 691, 1291, 1451, 2411, 3851, 4919, 5051, 5479, 5531, 5879, 6599, 7079, 7691, 8011, 8039, 11491, 13291, 14011, 15091, 15971, 16651, 17359, 18731, 19211, 19531, 20731, 22651, 23971, 24611, 25639, 25679, 26251, 32051, 32359, 32531, 32771, 32971, 35879, 37039, 37571, 38011, 38371
Offset: 1

Views

Author

Jianing Song, Jun 20 2025

Keywords

Comments

Primes p == 3 (mod 4) are precisely the rational primes in the ring of Gaussian integers.
5 is a quadratic residue of integers modulo p for p being a term of this sequence. (See A385165).

Examples

			5479 is a term since (2+-i)^125081 == 1 (mod 5479), and 125081 is divisible by neither 2 nor 3.
		

Crossrefs

Cf. A385165, A385179, A385219 (the actual multiplicative orders).
this sequence < A385169 < A385180 < A385167 < intersection of A122869 and A385168, where Ax < Ay means that Ax is a subsequence of Ay.
Also a subsequence of A385191.

Programs

  • PARI
    ord(p) = my(d = divisors((p+1)*znorder(Mod(5, p)))); for(i=1, #d, if(Mod([2, -1; 1, 2], p)^d[i] == 1, return(d[i]))) \\ for a prime p == 3 (mod 4), returns ord(2+-i, p)
    isA385188(p) = isprime(p) && p%4==3 && ord(p)%2 && ord(p)%3

A040105 Primes p such that x^4 = 5 has a solution mod p.

Original entry on oeis.org

2, 5, 11, 19, 31, 59, 71, 79, 101, 109, 131, 139, 149, 151, 179, 181, 191, 199, 211, 239, 251, 269, 271, 311, 331, 359, 379, 389, 401, 409, 419, 431, 439, 449, 461, 479, 491, 499, 521, 541, 569, 571, 599, 619
Offset: 1

Views

Author

Keywords

Comments

Union of 2, 5, A122869 (primes congruent to 11 or 19 modulo 20), and primes p == 1 (mod 4) such that 5^((p-1)/4) == 1 (mod p). - Jianing Song, Jun 20 2025

Crossrefs

Apart from 2 and 5, subsequence of A045468.
A385192 (which itself contains A122869) is a proper subsequence.

Programs

  • Magma
    [p: p in PrimesUpTo(800) | exists(t){x : x in ResidueClassRing(p) | x^4 eq 5}]; // Vincenzo Librandi, Sep 11 2012
    
  • Mathematica
    ok [p_]:=Reduce[Mod[x^4- 5, p] == 0, x, Integers] =!= False;  Select[Prime[Range[200]], ok] (* Vincenzo Librandi, Sep 11 2012 *)
  • PARI
    isA040105(p) = isprime(p) && (p==2 || p==5 || p%20==11 || p%20==19 || (p%4==1 && Mod(5,p)^((p-1)/4) == 1)) \\ Jianing Song, Jun 20 2025

A385191 Primes p == 3 (mod 4), p > 3 such that 2+-i are 24th powers modulo p.

Original entry on oeis.org

599, 691, 1039, 1291, 1451, 1759, 2411, 2879, 3079, 3491, 3851, 4519, 4639, 4919, 5051, 5479, 5519, 5531, 5639, 5879, 6011, 6079, 6599, 6719, 7079, 7691, 8011, 8039, 8171, 8731, 9439, 9839, 10799, 11159, 11239, 11411, 11491, 12239, 12799, 13291, 13679, 13759, 13879, 14011, 14639
Offset: 1

Views

Author

Jianing Song, Jun 20 2025

Keywords

Comments

Note that the primes congruent to 3 modulo 4 are precisely the rational primes in the ring of Gaussian integers.
Primes p == 3 (mod 4), p > 3 such that (2+-i)^((p^2-1)/24) == 1 (mod p). Note that p^2-1 is always divisible by 24 for primes p > 3.
Primes p = A002145(k) > 3 such that the multiplicative order of 2+-i modulo p (A385165(k)) divides (p^2-1)/24.
Primes p == 3 (mod 4), p > 3 such that [2,-1;1,2]^((p^2-1)/24) or [2,1;-1,2]^((p^2-1)/24) == I_2 (mod p).
Note that if (x+-y*i)^24 == 1+-i (mod p) for some integers x, y, then (x^2+y^2)^24 == 5 (mod p), so 5 must be a quadratic residue (in rational integers) modulo p. By definition, we have p == 11, 19 (mod 20).

Examples

			1759 is a term since (2+-i)^((1759^2-1)/24) = (-4)^((31^2-1)/96) = 1048576 == 1 (mod 31). Indeed, the solutions to x^24 == 2+i (mod 1759) are x == {441+580i, -43+860i, -292+683i, -251+779i, -635+872i, 736-648i} X {+-1, +-i} (mod 1759).
		

Crossrefs

Cf. A385165, A385190 (1+-i are 24th powers), A002145, A122869. A385188 is a subsequence.

Programs

  • PARI
    isA385191(p) = p>3 && isprime(p) && p%4==3 && Mod([2,-1;1,2],p)^((p^2-1)/24) == 1
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