%I #25 Mar 23 2024 07:26:12

%S 1,6,12,24,24,49,41,59,61,100,56,132,76,128,122,165,87,223,95,208,176,

%T 190,110,318,187,232,211,342,138,440,127,327,275,327,242,511,188,313,

%U 292,581,180,587,177,467,489,321,179,727,327,520,344,580,193,713,408,640,410,442,222,1055,288

%N Number of solutions to n == xy (mod z) == yz (mod x) == zx (mod y) with 0 < x < y < z.

%C 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.

%C Following the linked solution by Silvia Fernández, it can be shown that all solutions satisfy n < x <= 3n^2 - n, x < y <= 2nx - n, and y < z <= xy - n. Contrary to the above comment, there are solutions satisfying x > 2n^2 + n. The first example is given by (x,y,z) = (434,630,826) when n = 14. - _Robin Visser_, Dec 18 2023

%H M. F. Possner, <a href="http://www.jstor.org/stable/3647915">Problem 11021</a>, Amer. Math. Monthly, 110 (2003), p. 542.

%H Donald Knuth, Silvia Fernández and Gerry Myerson, <a href="http://www.jstor.org/stable/30037459">A Modular Triple: 11021</a>, Amer. Math. Monthly, 112 (2005), p. 279.

%e 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).

%o (Sage)

%o def a(n):

%o ans = 0

%o for x in range(n+1, 3*n^2-n+1):

%o for y in range(x+1, 2*n*x-n+1):

%o for z in Integer(x*y-n).divisors():

%o if (z > y) and ((x*y)%z)==n and ((y*z)%x)==n and ((z*x)%y)==n:

%o ans += 1

%o return ans # _Robin Visser_, Dec 12 2023

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

%K nonn

%O 1,2

%A _T. D. Noe_, May 06 2004, revised May 13 2004

%E Corrected and a(33)-a(46) added by _Robin Visser_, Dec 18 2023

%E a(47)-a(61) from _Robin Visser_, Mar 18 2024