A236183 Primes p such that for all primes q dividing p-1 every residue mod p is the sum of two q-th powers mod p.

2, 3, 5, 13, 17, 19, 37, 73, 97, 101, 109, 151, 163, 181, 193, 197, 211, 241, 251, 257, 271, 281, 337, 379, 397, 401, 421, 433, 449, 487, 491, 541, 577, 601, 631, 641, 661, 673, 701, 727, 751, 757, 769, 811, 881, 883, 991, 1009

Offset

1,1

Comments

Alternative definition: increasing list of primes p such that for ALL primes q, every residue mod p is the sum of two q-th powers mod p (if q does not divides p-1 then every residue mod p is a q-th power, so only the case q divides p-1 is not trivial).

Related to the conjecture:

For every prime q there are only finitely many primes p such that not every residue mod p is the sum of two q-th powers mod p.

Links

Charles R Greathouse IV, Table of n, a(n) for n = 1..400

Example

p=7 is not in the list because q=3 divides p-1 and the only residues mod 7 which are the sum of two cubic residues mod 7 are: 0,1,-1,2,-2.
To check that p=13 is in the sequence, since 2 and 3 are the only primes dividing p-1=12, we only need to see that every residue mod 13 is both the sum of two quadratic residues mod 13 and the sum of two cubic residues mod 13.

Prog

(Sage)
for p in prime_range(a, b):
    c=1
    C=GF(p)
    u=combinations_with_replacement(C, 2)
    v=[x for x in u]
    for q in prime_divisors(p-1):
        w=(k[0]^q+k[1]^q for k in v)
        s=set(w)
        l=len(s)
        if l!=p:
            c=0
            break
    if c==1:
        print(p)
(PARI) is(p)=if(!isprime(p), return(0)); my(f=factor(p-1)[, 1], v, u); for(i=1, #f, u=vector(p); v=vector(p, j, j^f[i]%p); for(j=1, p, for(k=j, p, u[(v[j]+v[k])%p+1]=1)); if(!vecmin(u), return(0))); 1 \\ Charles R Greathouse IV, Jan 23 2014

Crossrefs

Sequence in context: A259188 A173971 A095083 * A093077 A360418 A249016

Adjacent sequences: A236180 A236181 A236182 * A236184 A236185 A236186

Keyword

nonn

Author

Esteban Arreaga Ambéliz, Jan 19 2014

Status

approved