A324599 Irregular triangle with the representative solutions of the Diophantine equation x^2 - 5 congruent to 0 modulo N(n), with N(n) = A089270(n), for n >= 1.

0, 0, 4, 7, 9, 10, 11, 18, 6, 25, 13, 28, 15, 40, 8, 51, 26, 35, 17, 54, 20, 59, 19, 70, 10, 85, 45, 56, 21, 88, 48, 73, 23, 108, 12, 127, 40, 105, 68, 81, 55, 96, 25, 130, 30, 149, 27, 154, 14, 177, 76, 123, 95, 110, 29, 180, 48, 161, 65, 146

Offset: 1

Author

Wolfdieter Lang, Jul 08 2019

Keywords

• nonn
• tabf

Comments

The length of row n is 1 for n = 1 and n = 2, and for n >= 3 it is 2^{r1 + r4} with the number r1 and r4 of distinct primes congruent to 1 and 4 modulo 5, respectively, in the prime number factorization of N(n). E.g., n = 29, N = 209 = 11*19, has r1 = 1 and r4 = 1, with four solutions.

For N(1) = 1 every integer solves this Diophantine equation, and the representative solution is 0.

For N(2) = 5 there is only one representative solution, namely 0.

For n >= 3 the solutions come in a nonnegative power of 2 pairs, each of the type (x1, x2) with x2 = N - x1.

See the link in A089270 to the W. Lang paper, section 3, and Table 7.

Examples

			The irregular triangle T(n, k) begins (pairs (x, N - x) in brackets):
n,    N \ k   1   2     3   4  ...
----------------------------------
1,    1:      0
2,    5:      0
3,   11:     (4   7)
4,   19:     (9  10)
5,   29:    (11  18)
6,   31:     (6  25)
7,   41:    (13  28)
8,   55:    (15  40)
9,   59:     (8  51)
10,  61:    (26  35)
11,  71:    (17  54)
12,  79:    (20  59)
13,  89:    (19  70)
14,  95:    (10  85)
15, 101:    (45  56)
16, 109:    (21  88)
17, 121:    (48  73)
18, 131:    (23 108)
19, 139:    (12 127)
20, 145:    (40 105)
....
29, 209:    (29 180)  (48 161)
...
41, 319:    (18 301)  (40 279)
...
43, 341:    (37 304) (161 180)
...
59, 451:    (95 356) (136 315)
		

Crossrefs

Cf. A089270, A324598 (x^2 + x - 1 == 0 (mod N)).