A322781 Numbers of the form p*q where p, q are distinct primes congruent to 1 mod 4 such that Legendre(p/q) = -1.

65, 85, 185, 265, 365, 481, 485, 493, 533, 565, 629, 685, 697, 785, 865, 949, 965, 985, 1037, 1073, 1157, 1165, 1189, 1241, 1261, 1285, 1385, 1417, 1465, 1565, 1585, 1649, 1685, 1765, 1769, 1781, 1853, 1865, 1921, 1937, 1985, 2117, 2165, 2173, 2257, 2285, 2509, 2561, 2581, 2785, 2813, 2885, 2929, 2941

Offset: 1

Author

N. J. A. Sloane, Jan 11 2019

Keywords

• nonn

Comments

If k is a term, the Pell equation x^2 - k*y^2 = -1 has a solution [Dirichlet, Newman (1977)]. This is only a sufficient condition, there are many other solutions, see A031396.

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000
• Morris Newman, A note on an equation related to the Pell equation, The American Mathematical Monthly 84.5 (1977): 365-366.

Crossrefs

Cf. A002144, A031396, A323271, A323272.

Programs

PARI

isok(n) = my (f=factor(n)); omega(f)==2 && bigomega(f)==2 && f[1,1]%4==1 && f[2,1]%4==1 && kronecker(f[1,1], f[2,1])==-1 \\ Rémy Sigrist, Jan 11 2019

Python

from sympy.ntheory import legendre_symbol, factorint
A322781_list, k = [], 1
while len(A322781_list) < 10000:
    fk, fv = zip(*list(factorint(4*k+1).items()))
    if sum(fv) == len(fk) == 2 and fk[0] % 4 == fk[1] % 4 == 1 and legendre_symbol(fk[0],fk[1]) == -1:
            A322781_list.append(4*k+1)
    k += 1 # Chai Wah Wu, Jan 11 2019