OFFSET
1,1
COMMENTS
Solution to a Diophantine equation in finite fields Z/n. Historical reminder: Sophie Germain's work led to the breaking of Fermat's Last Theorem into two cases: x^p + y^p = z^p has no integer solutions for which x, y and z are relatively prime to p, i.e., in which none of x, y and z are divisible by p, and then x^p + y^p = z^p has no integer solutions for which one of the three numbers is divisible by p.
This result was presented by Legendre in an 1823 paper to the French Academy of Sciences and included in a supplement to his second edition of Theorie des Nombres, with a footnote crediting the result to Sophie Germain. Sophie Germain's Theorem introduced an auxiliary prime n satisfying the two conditions: x^p + y^p + z^p == 0 (mod n) implies that x == 0 (mod n), or y == 0 (mod n), or z == 0 (mod n), and x^p == p (mod n) is impossible for any value of x. Then Case I of Fermat's Last Theorem is true for p. This sequence gives solutions for each prime number p, and n = 2p + 1.
REFERENCES
René Schoof, "Wiles' proof of the Taniyama-Weil conjecture for semi-stable elliptic curves over Q", Chap. 14 in 'Où en sont les Mathématiques ?' Soc. Math. de France (SMF), Vuibert, Paris 2002.
LINKS
Robin Visser, Table of n, a(n) for n = 1..1000
C. K. Caldwell, The Prime Glossary, Fermat's Last Theorem
Andrea Del Centina, Letters of Sophie Germain preserved in Florence, Historia Mathematica, Vol. 32 (2005), 60-75.
Andrea Del Centina, Unpublished manuscripts of Sophie Germain and a revaluation of her work on Fermat's Last Theorem, Arch. Hist. Exact Sci., Vol 62 (2008), 349-392.
A. M. Legendre, Recherches sur quelques objets d'analyse indéterminée et particulièrement sur le théorème de Fermat, Mem. Acad. Sci. Inst. France 6 (1823), 1-60.
J. H. Sampson, Sophie Germain and the theory of numbers, Arch. Hist. Exact Sci. 41 (1990), 157-161.
EXAMPLE
We consider the case p = 1, n = 3. We have 5 solutions mod 3: (0,1,2), (0,2,1), (1,1,1), (1,2,0), (2,2,2).
With p = 2, n = 5, we have 12 solutions mod 5: (0,1,2), (0,1,3), (0,2,1), (0,2,4), (0,3,1), (0,3,4), (0,4,2), (0,4,3), (1,2,0), (1,3,0), (2,4,0), (3,4,0),
With p = 3, n = 7, we have 27 solutions mod 7: (0,1,3), (0,1,5), (0,1,6), (0,2,3), (0,2,5), (0,2,6), (0,3,1), (0,3,2), (0,3,4), (0,4,3), (0,4,5), (0,4,6), (0,5,1), (0,5,2), (0,5,4), (0,6,1), (0,6,2), (0,6,4), (1,3,0), (1,5,0), (1,6,0), (2,3,0), (2,5,0), (2,6,0), (3,4,0), (4,5,0), (4,6,0).
PROG
(Sage)
p = 1
while (p < 1000):
n, ans = 2*p + 1, 0
numz = [0 for i in range(n)]
for i in range(n): numz[power_mod(i, p, n)] += 1
for y in range(1, n):
for x in range(y+1):
ans += numz[(-power_mod(x, p, n)-power_mod(y, p, n))%n]
print(ans)
p = p.next_prime()
while(not (2*p+1).is_prime()): p = p.next_prime() # Robin Visser, Jul 29 2023
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Feb 02 2010
EXTENSIONS
More terms from Robin Visser, Jul 29 2023
STATUS
approved