1,2

Possner (see also Knuth) asks for solutions when n=2. There is at least one solution for all positive n: (x,y,z) = (n+1, n^2+2n, n^3+3n^2+n). All solutions appear to be in the polytope n < x <= 2n^2+n, x < y <= 2n^3+2n^2-n, y < z <= n^5+2n^4+2n^3+n^2-n. Many solutions, especially for prime n, are such that n divides x, y and z. See A094595.

M. F. Possner, Problem 11021, Amer. Math. Monthly, 110 (2003), p. 542

D. E. Knuth, Solution 11021, Amer. Math. Monthly, 112 (2005), p. 279

a(2) = 6 because there are 6 solutions: (x,y,z) = (3, 8, 22), (3, 10, 14), (4, 5, 18), (4, 6, 11), (6, 14, 82) and (6, 22, 26).

Table[cnt=0; Do[d=Divisors[x*y-n]; Do[z=d[[i]]; If[z>y, If[Mod[x*z, y]==n && Mod[y*z, x]==n, cnt++; ]], {i, Length[d]}], {x, n+1, n(2n+1)}, {y, x+1, n(2n^2+2n-1)}]; cnt, {n, 10}]

Cf. A094595 (number of solutions to 1 = nxy (mod z) = nyz (mod x) = nzx (mod y) with 0<x<y<z).

Sequence in context: A005694 A081512 A096387 this_sequence A074902 A096366 A061822

Adjacent sequences: A094182 A094183 A094184 this_sequence A094186 A094187 A094188

more,nonn

T. D. Noe (noe(AT)sspectra.com), May 06 200, revised May 13 2004