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A208296
Smallest positive nontrivial odd solution of the congruence x^2 == 1 (mod A001748(n+2)), n >= 1.
4
11, 13, 23, 25, 35, 37, 47, 59, 61, 73, 83, 85, 95, 107, 119, 121, 133, 143, 145, 157, 167, 179, 193, 203, 205, 215, 217, 227, 253, 263, 275, 277, 299, 301, 313, 325, 335, 347, 359, 361, 383, 385, 395, 397, 421, 445, 455, 457, 467, 479, 481, 503, 515
OFFSET
1,1
COMMENTS
The trivial solutions of the congruence x^2 == 1 (mod 3*prime(n+2)), n>=1, with the primes prime(n+2) = A000040(n+2) have positive representatives 1 and 3*prime(n+2)-1. There are all-together four incongruent solutions due to a general theorem (see, e.g., the Hardy-Wright reference, Theorem 122, p. 96, and also A060594) and the fact that the number of incongruent solutions of this congruence with odd prime modulus p is two, namely with positive representative p and p-1 (see, e.g., Hardy-Wright, Theorem 109, p. 85). a(n) is the smallest positive odd representative >1 which solves this congruence. The other nontrivial even representative solving this congruence is 3*prime(n+2) - a(n), i.e. 4, 8, 10, 14, 16, 20, ... See 2*A207336.
a(n) solves also the congruence x^2 == 1 (Modd A001748(n+2)), n>=1. For Modd n (not to be confused with mod n) see a comment on A203571. This follows from floor(a(n)^2/3*prime(n+2)) being even, in fact it is 8*A024699(n) (see a comment there), hence a(n)^2 (Modd 3*prime(n+2)) = a(n)^2 (mod 3*prime(n+2)) = 1. For those multiplicative groups Modd 3*p with p an odd prime which are cyclic (this is not possible in the mod case, see A033949), a(n) is the representative of the only other nontrivial solution of this congruence. The representative of the trivial solution is 1 (-1 belongs to the same Modd class). (The conjecture stated here earlier is wrong, that is, the multiplicative group Modd (91=7*13) is non-cyclic. It may still be true for 3*p. - Wolfdieter Lang, Mar 15 2012)
REFERENCES
H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 5th ed., Clarendon Press, Oxford, 2003.
FORMULA
a(n) = sqrt(8*T(K(n))+1), with the triangular numbers T = A000217, and K(n) = prime(n+2)-1 if the prime prime(n+2) is of the form 6*k+1, i.e., from A002476, and K(n) = prime(n+2) if prime(n+2) is of the form 6*k-1, i.e. from A007528.
a(n)^2 == 1 (mod A001748(n+2)), n >= 1.
a(n)^2 == 1 (Modd A001748(n+2)), n >= 1.
EXAMPLE
a(3)=23 because prime(5)=11=A007528(2), hence K(3)=11 and sqrt(8*T(11)+1)=sqrt(8*66+1)= 23. 23^2 = 529 == 1 (Modd 33), because floor(529/33)=16=8*A024699(3) is even, and 529 == 1 (mod 33).
a(4)=25 because prime(6)=13=A002476(2), hence K(4)=12 and sqrt(8*T(12)+1)=sqrt(8*78+1)=25. 25^2 = 625 == 1 (Modd 39), because floor(625/39)=16=8*A024699(4) is even, and 625 == 1 (mod 39).
MATHEMATICA
Table[SelectFirst[Solve[x^2==1 && x !=1, x, Modulus->3*Prime[n+2]][[All, 1, 2]], OddQ], {n, 53}] (* Jon Maiga, Sep 28 2019 *)
KEYWORD
nonn
AUTHOR
Wolfdieter Lang, Mar 14 2012
STATUS
approved