|
| |
|
|
A212353
|
|
a(n) is the smallest positive solutions of the congruence x^2 + (x+1)^2 == 0 (mod prime), where prime = A002144(n) (Pythagorean primes).
|
|
1
|
|
|
|
1, 2, 6, 8, 15, 4, 11, 5, 13, 27, 37, 45, 16, 7, 18, 52, 64, 46, 9, 40, 91, 53, 44, 88, 120, 93, 108, 26, 77, 12, 101, 94, 106, 155, 134, 57, 31, 190, 71, 14, 89, 33, 54, 206, 150, 117, 244, 219, 241, 276, 38, 62, 17, 211, 243, 74, 277, 307, 325, 67, 306, 176, 43
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
There are at most two incongruent solutions of this congruence due to the degree. The fact that there are precisely two such solutions for each prime of the form 4*k+1 (see A002144) is due to the reduction of this problem to one of quadratic residues, namely to X^2 == -1 (mod 2p), with p a prime (see the Nagell reference, given in A210848, pp. 132-3, especially theorem 77), adapted to the quadratic form f(x) = 2*x^2 + 2*x + 1, with discriminant D=-4. This congruence with composite modulus has exactly two incongruent solutions because X^2 == -1 (mod 2) has only the solution +1 modulo 2 (odd numbers), and X^2 == -1 (mod p) has (at least one) solution if the Legendre symbol (-1/p) = +1 (i.e., if -1 is a quadratic residue modulo p). Now (-1/p) = (-1)^(p-1)/2 (see, e.g., the Niven-Zuckerman-Montgomery reference given in A001844, Theorem 3.2 (1), p. 132). Hence there is a solution modulo p iff p == 1 (mod 4). Call the smallest positive one X0, with 0 < X0 < p-1. Then one also has the incongruent solution X1 := p-X0. This implies that there are precisely two incongruent solution of the original congruence modulo 2*p for each 1 (mod 4) prime (see, e.g., Nagell's book, pp. 83-4, Theorem 46). If u is a solution for p = A002144(n) (the existence of u has just been proved) then also the companion v := p-1-u satisfies this congruence, and v is incongruent to u modulo p.
Note that x^2 + (x+1)^2 = 4*T(x) + 1, with the triangular numbers A000217.
The primes with x^2 +(x+1)^2 = prime (necessarily from A002144) are found under A027862. The corresponding x values are found under A027861. These x values explain the positions n' where a(n') is smaller than a(n'-1) (for n'>=6): determine k with x=A027861(k), and then n' from A027862(k) = A002144(n'). Note that a(n') = x for such values n'. E.g., n'=6 with a(6)=4: x=4=A027861(3), p=41=A027862(3) = A002144(6). These values n' are n' = 1, 2, 6, 8, 14, 19, 30, ...
All positive solutions of this congruence are provided by the two sequences with entries u(n,k) = a(n) + k*A002144(n) and v(n,k) = A212354(n) + k*A002144(n), n >= 1, k >= 0. For the cases p = 5, 13 and 17 see A047219, A212160 and A212161, respectively, where the even-indexed numbers are the u(n,k) and the odd-indexed ones the v(n,k) (bisection).
2*a(n) + 1 = A206549(n), the smallest positive nontrivial solution of X^2 == +1 (Modd A002144(n)). For the next larger solution 2*A212354(n) + 1 >= p, hence it does not belong to the restricted residue system Modd A002144(n).
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) is the smaller of the two smallest positive incongruent solutions of the congruence x^2 + (x+1)^2 = 2*x^2 + 2*x + 1 == 0 (mod A002144(n)), where A002144 lists the primes
1 modulo 4 (primes of the form 4*k+1). For the proof of the existence of a(n) see a comment above. The next larger incongruent companion solution is A212354(n), n >= 1.
|
|
|
EXAMPLE
|
n=1: a(1)=1 because 1^2 + 2^2 = 5 == 0 (mod 5). The companion solution is (5-1) - 1 = 3 = A212354(1).
n=3: a(3)=6 because 6^2 + 7^2 = 85 = 5*17 == 0 (mod 17). The companion is (17-1) - 6 = 10 = A212354(3).
n=14: a(14)=7 because p=A002144(14) = 113 = A027862(5), and 49^2 + 50^2 = 113. The companion is (113-1) - 7 = 105 = A212354(14).
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
