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A324599
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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.
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1
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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
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OFFSET
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1,3
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COMMENTS
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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.
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LINKS
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EXAMPLE
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The irregular triangle T(n, k) begins (pairs (x, N - x) in brackets):
n, N \ k 1 2 3 4 ...
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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)
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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