A371670 Largest prime number p such that x^n + y^n + z^n mod p does not take all values on Z/pZ.

5, 11, 31, 43, 41, 37, 61, 89, 229, 79, 127, 61, 257, 137, 397, 419, 521, 337, 463, 277, 577, 251, 599, 541, 617, 349, 661, 373, 769, 727, 953, 631, 1117, 593, 761, 1483, 761, 739, 1597, 1033, 1409, 1171, 1289, 1693, 2113, 883, 1301, 1327, 2861, 1697, 2269, 1871, 2633, 1483, 2089, 2243, 2221, 1709, 1861, 2143

Offset: 4

Author

Robin Visser, Apr 02 2024

Keywords

• nonn

Comments

a(n) is well-defined for all n >= 4, following a similar discussion given in A355920. Indeed for any integer k, by fibering the surface C : x^n + y^n + z^n == k (mod p) into curves and applying the Hasse-Weil bound, we obtain that the number of points N on C must satisfy |N - p(p+1)| <= 2*g*p*sqrt(p), where g = (n-1)(n-2)/2 is the genus of the Fermat curve x^n + y^n = 1. Thus, N is nonzero if p+1 > (n-1)(n-2)*sqrt(p). In particular, x^n + y^n + z^n mod p takes all values on Z/pZ for all primes p > n^4.

a(n) <= A355920(n). - Jason Yuen, May 30 2024

Examples

			For n = 4, the equation x^4 + y^4 + z^4 == 4 (mod 5) does not have a solution, but x^4 + y^4 + z^4 == k (mod p) does have a solution for any integer k and prime p greater than 5, thus a(4) = 5.
For n = 5, the equation x^5 + y^5 + z^5 == 4 (mod 11) does not have a solution, but x^5 + y^5 + z^5 == k (mod p) does have a solution for any integer k and prime p greater than 11, thus a(5) = 11.
		

Links

• S. Lang and A. Weil, Number of Points of Varieties in Finite Fields, Amer. J. Math., vol. 76, no. 4, 1954, pp. 819-27.
• MathOverflow, Does the expression x^4+y^4 take on all values in Z/pZ?

Crossrefs

Cf. A355920, A368055.

Programs

Python

from itertools import combinations_with_replacement
from sympy import prevprime
def A371670(n):
    p = n**4
    while (p:=prevprime(p)):
        pset = set(q:=tuple(pow(x,n,p) for x in range(p)))
        if not all(any((k-a[0]-a[1])%p in pset for a in combinations_with_replacement(q,2)) for k in range(p)):
            return p # Chai Wah Wu, Apr 04 2024

SageMath

def a(n):
    p = Integer(n^4).previous_prime()
    while True:
        nth_powers = set([power_mod(x, n, p) for x in range(p)])
        for k in range(p):
            for xn, yn in ((x,y) for x in nth_powers for y in nth_powers):
                if (k-xn-yn)%p in nth_powers: break
            else: return p
        p = p.previous_prime()

Extensions

a(36)-a(63) from Jason Yuen, May 30 2024