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   Author  Topic: An Integer Triplet Puzzle  (Read 1050 times)
K Sengupta
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An Integer Triplet Puzzle  
« on: Feb 16th, 2007, 7:43am »		 Quote Quote Modify Modify
Determine all possible triplets of positive integers  (x, y, z) satisfying:
 
(y3 + 27)/(xy - 9) = z.
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jollytall
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Re: An Integer Triplet Puzzle  
« Reply #1 on: Feb 16th, 2007, 1:19pm »		 Quote Quote Modify Modify
Let's go systematically along y. So far:
 
y=0. z negative no solution
y=1. x-9 divides 28, so x=10, 11, 13, 16, 23
y=2. 2x-9 divides 35, so x=5, 7, 8.
y=3. 3x-9 divides 54. x-3 divides 18. x=4, 5, 6, 9, 12
y=4. 4x-9 divides 91. x=4
y=5. 5x-9 divides 152. x=2
y=6. 6x-9 divides 243, 2x-3 divides 81. x=2, 3, 6, 15, 42
y=7. 7x-9 divides 370=2*5*37. x=2
y=8. 8x-9 divides 539=7*7*11. x=2
y=9. 9x-9 divides 756, x-1 divides 84. x=2, 3, 42, 85
y=10. 10x-9 divides 1027=13*79. x=1
y=11. 11x-9 divides 1358=2*7*97. x=1.
y=12. 12x-9 divides 1755, 4x-3 divides 585=3*3*5*13. x=1, 2, 3, 4, 12, 17, 30, 147
y=13. 13x-9 divides 2224=2*2*2*2*139. x=1
y=14. 14x-9 divides 2771=17*163. No solution.
y=15. 15x-9 divides 3402. 5x-3 divides 1134=2*3*3*3*3*7. x=1, 2, 6
y=16. 16x-9 divides 4123=7*19*31. x=1.
y=17. 17x-9 divides 4940=2*2*5*13*19. No solution?
y=18. 18x-9 divides 5859, 2x-1 divides 651=3*7*31. x=2, 4, 16, 326
y=19. 19x-9 divides 6886=2*11*313. No solution.
y=20. 20x-9 divides 8027=23*349. No solution.
 
« Last Edit: Feb 16th, 2007, 1:27pm by jollytall » IP Logged
Eigenray
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Re: An Integer Triplet Puzzle  
« Reply #2 on: Mar 5th, 2007, 5:45pm »		 Quote Quote Modify Modify
The values of y<40,000,000 for which y3+27 has a divisor of the form xy-9 are
 
y=1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 21, 22, 23, 25, 27, 30, 33, 36, 37, 42, 45, 51, 57, 63, 72, 90, 93, 99, 111, 114, 117, 129, 135, 144, 147, 162, 171, 186, 198, 207, 225, 261, 297, 309, 315, 387, 477, 549, 684, 693, 765, 774, 855, 900, 981, 1233, 1341, 1773, 1782, 1791, 1845, 1872, 1962, 2205, 2682, 2817, 2844, 3312, 3330, 3789, 5607, 5742, 6597, 7065, 7470, 8757, 8766, 9297, 9567, 10062, 10161, 10872, 11457, 12051, 12222, 14211, 17406, 18702, 19071, 20097, 23607, 29871, 32517, 33597, 42462, 45405, 48771, 80361, 93006, 134010, 137277, 181557, 270126, 535851,
 
but I don't see the pattern.  One thing to note is that
 
(y3+27)x3 = (xy-9)(x2y2+9xy+81) + 27(x3+27),
 
so if xy-9 divides y3+27, it also divides 27(x3+27).
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Re: An Integer Triplet Puzzle  
« Reply #3 on: Jun 23rd, 2007, 8:00pm »		 Quote Quote Modify Modify
As I suspected, there are only finitely many solutions, and the above values of y give all of them.
 
Suppose xy - 9 | y3+27.  Then mod xy - 9, we have
 
0 (y3 + 27)*x = y3x + 27x 9y2 + 27x,
 
so we can write
 
(*)  9y2 + 27x = (xy - 9)*k
 
for some integer k > 0, and solving for y, we find that the discriminant
 
D = (kx)2 - 4*9*(27x+9k)
 
is a perfect square.  Since clearly D < (kx)2, we must have D (kx-1)2, which can be rearranged to give
 
2kx - 4.34(k+3x)   1,
 
or
 
(2k - 4.35)(x - 2.34) 8.39+1.
 
Putting k=x=r above gives equality for r ~ 648.001, and so we can conclude that at least one of k, x must be 648.
 
We can also rewrite (*) as
 
9(y2 + k) = x(ky - 27),
 
and therefore, mod (ky-27), we have
 
0 = 9(y2+k)*k2 38 + 9k3.
 
So for each value of k 648, we can see if 38+9k3 has any divisors of the form ky-27, and get possible values of y.  And if k > 648, then x 648, so for each such x, we can see if 27(x3+27) has any divisors of the form xy-9, and get the remaining possible values of y.
 
Going through these values to see for which y3+27 has a divisor of the form xy-9, the set of all solutions, in the format (y, x), is therefore:
 
(1, 10), (1, 11), (1, 13), (1, 16), (1, 23), (1, 37), (2, 5), (2, 7), (2, 8), (2, 22), (3, 4), (3, 5), (3, 6), (3, 9), (3, 12), (3, 21), (4, 4), (4, 25), (5, 2), (5, 17), (6, 2), (6, 3), (6, 6), (6, 15), (6, 42), (7, 2), (8, 2), (9, 2), (9, 3), (9, 4), (9, 5), (9, 7), (9, 8), (9, 13), (9, 15), (9, 22), (9, 29), (9, 43), (9, 85), (10, 1), (11, 1), (12, 1), (12, 2), (12, 3), (12, 4), (12, 12), (12, 17), (12, 30), (12, 147), (13, 1), (15, 1), (15, 2), (15, 6), (15, 9), (15, 33), (15, 114), (16, 1), (17, 5), (18, 1), (18, 2), (18, 4), (18, 11), (18, 16), (18, 47), (18, 109), (18, 326), (21, 1), (21, 3), (21, 25), (21, 111), (22, 2), (23, 1), (25, 4), (27, 1), (27, 2), (27, 49), (27, 122), (30, 1), (30, 8), (30, 12), (30, 129), (33, 7), (33, 15), (36, 1), (36, 2), (36, 5), (36, 10), (36, 23), (36, 62), (36, 100), (36, 1297), (37, 1), (42, 6), (42, 22), (45, 1), (45, 5), (45, 169), (45, 1013), (51, 1), (51, 186), (57, 2), (57, 93), (63, 1), (63, 361), (72, 1), (72, 2), (72, 10), (72, 22), (72, 247), (72, 1037), (90, 4), (90, 121), (93, 1), (93, 5), (93, 57), (93, 309), (99, 2), (99, 65), (111, 16), (111, 21), (114, 15), (114, 23), (117, 1), (117, 1369), (129, 13), (129, 30), (135, 1), (135, 434), (144, 7), (144, 340), (147, 12), (147, 37), (162, 2), (162, 250), (171, 142), (186, 11), (186, 51), (198, 1), (198, 64), (207, 4), (207, 5), (207, 17), (207, 157), (225, 16), (225, 205), (225, 415), (261, 1), (261, 8), (261, 811), (297, 19), (297, 47), (309, 10), (309, 93), (315, 14), (315, 1481), (387, 1), (477, 2), (477, 7), (477, 11), (477, 205), (549, 43), (684, 28), (684, 6409), (693, 896), (765, 1), (774, 1403), (855, 140), (900, 427), (981, 121), (1233, 37), (1233, 100), (1341, 12317), (1773, 515), (1782, 14), (1782, 382), (1791, 4045), (1845, 32), (1845, 173), (1872, 22), (1962, 17), (2205, 89), (2682, 142), (2817, 55), (2844, 400), (3312, 30053), (3330, 15046), (3789, 29), (3789, 392), (5607, 1442), (5742, 380), (6597, 59617), (7065, 28), (7065, 757), (7470, 44), (8757, 157), (8766, 155), (9297, 214), (9567, 229), (10062, 119), (10161, 118), (10872, 1997), (11457, 109), (12051, 340), (12222, 347), (14211, 100), (17406, 95), (18702, 599), (19071, 613), (20097, 1447), (23607, 784), (29871, 88), (32517, 1117), (33597, 4957), (42462, 1487), (45405, 41), (48771, 85), (80361, 2893), (93006, 83), (134010, 9886), (137277, 5002), (181557, 82), (270126, 9923), (535851, 19765).
 
Is there a method without so much brute force?
« Last Edit: Jun 23rd, 2007, 8:12pm by Eigenray » IP Logged
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