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A094595
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Number of solutions to 1 == nxy (mod z) == nyz (mod x) == nzx (mod y) with 0 < x < y < z.
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2
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1, 2, 6, 5, 17, 3, 31, 7, 23, 5, 47, 5, 60, 15, 20, 12, 78, 12, 78, 10, 35, 18, 91, 9, 74, 19, 50, 9, 120, 7, 110, 38, 56, 30, 65, 9, 171, 41, 67, 12, 169, 16, 143, 36, 55, 17, 162, 12, 143, 19, 55, 28, 171, 13, 113, 23, 71, 33, 201, 6, 265, 50, 59, 45
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OFFSET
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1,2
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COMMENTS
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Prime values of n yield many more solutions than composite values. If (x,y,z) is a solution, then (nx,ny,nz) is a solution of the equations in A094185. All solutions appear to be in the polytope n < x <= 2n+1, x < y <= 2n^2+2n-1, y < z <= n^4+2n^3+2n^2+n-1.
There are at least two solutions for all n > 1, given by (x,y,z) = (n+1, n^2+n+1, n^4+2n^3+2n^2+n-1) and (x,y,z) = (n+1, 2n^2+2n-1, 2n^2+2n+1). Following the linked solution by Silvia Fernández, it can be shown that all solutions satisfy 1 < x <= 3n - 1, x < y <= 2nx - 1, and y < z <= nxy - 1. Contrary to the above comment, there are solutions satisfying x > 2n+1. The first such example is given by (x,y,z) = (31,45,59) when n = 14. - Robin Visser, Dec 18 2023
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LINKS
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Donald Knuth, Silvia Fernández and Gerry Myerson, A Modular Triple: 11021, Amer. Math. Monthly, 112 (2005), p. 279.
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EXAMPLE
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a(2) = 2 because there are 2 solutions: (x,y,z) = (3, 7, 41) and (3, 11, 13).
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MATHEMATICA
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Table[cnt=0; Do[d=Divisors[n*x*y-1]; Do[z=d[[i]]; If[z>y && Mod[n*x*z, y]==1 && Mod[n*y*z, x]==1, cnt++ ], {i, Length[d]}], {x, 3n-1}, {y, x+1, 2n*x-1}]; cnt, {n, 64}]
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PROG
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(Sage)
def a(n):
ans = 0
for x in range(2, 3*n):
for y in range(x+1, 2*n*x):
for z in Integer(n*x*y-1).divisors():
if (z > y) and ((n*y*z)%x)==1 and ((n*z*x)%y)==1:
ans += 1
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CROSSREFS
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Cf. A094185 (number of solutions to n = xy (mod z) = yz (mod x) = zx (mod y) with 0<x<y<z).
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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