A190728 Least m>1 such that m^3 mod n^2 is 1.

5, 4, 17, 26, 13, 18, 65, 28, 101, 122, 49, 22, 165, 76, 257, 290, 109, 68, 401, 67, 485, 530, 193, 626, 529, 244, 177, 842, 301, 439, 1025, 364, 1157, 226, 433, 581, 429, 22, 1601, 1682, 361, 423, 1937, 676, 2117, 2210, 769, 1047, 2501, 868, 529, 2810, 973, 3026, 961, 292, 3365, 3482, 1201, 1660

Offset

2,1

Comments

For n = 2, 4, 5 and 8 there are the obvious solutions n^2 + 1, but these are not minimal for n = 3, 6, 7 and 9.

Which other n have least m less than n^2 + 1?

There will be a solution less than n^2 + 1 when n is divisible by 3 or by any prime == 1 (mod 3). - Franklin T. Adams-Watters, May 17 2011.

Links

Table of n, a(n) for n=2..61.

Example

The obvious case for 4 is 4^2 +1 = 17 and 17^3=4913 == 1 (mod 16).
For 6 the least m^3 is not the obvious m=36^2+1=37 but m=13 to give 13^3=2197 == 1 (mod 49).

Prog

(PARI) a(n)=for(k=2, n^2+1, if(Mod(k, n^2)^3==1, return(k)))

Crossrefs

Sequence in context: A281843 A282225 A282202 * A344435 A100791 A056883

Adjacent sequences: A190725 A190726 A190727 * A190729 A190730 A190731

Keyword

nonn

Author

J. M. Bergot, May 17 2011

Extensions

Corrected, extended, and edited by Franklin T. Adams-Watters, May 17 2011

Status

approved