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        riddles >> medium >> three square numbers
        
(Message started by: Christine on Aug 15th, 2013, 1:00pm)
Title: three square numbers
Post by Christine on Aug 15th, 2013, 1:00pm
To find three numbers (x, y, z) such that

x + (x*y*z)
y + (x*y*z)
z + (x*y*z)
are all squares
one possible solution:
x = 1/9, y = 9/9 and z = 40/9

any other solution?
Please show your analytical solution
Title: Re: three square numbers
Post by towr on Aug 18th, 2013, 7:08am
Plenty of other solutions, here's a few with numerator and divisor under 100:
1/24 2/24 12/24
1/8 4/8 34/8
1/48 8/48 96/48
1/9 9/9 40/9
1/12 12/12 52/12
1/15 17/15 30/15
1/4 20/4 64/4
1/32 34/32 64/32
1/23 46/23 49/23
2/8 13/8 20/8
2/8 17/8 36/8
2/40 20/40 41/40
3/12 32/12 80/12
4/24 6/24 25/24
4/13 13/13 68/13
4/12 16/12 39/12
4/44 16/44 55/44
4/72 18/72 73/72
4/88 22/88 89/88
4/56 26/56 35/56
4/56 61/56 70/56
5/12 12/12 68/12
5/44 36/44 44/44
6/24 20/24 29/24
7/28 16/28 32/28
8/32 36/32 41/32
9/72 49/72 72/72
9/36 52/36 64/36
9/56 56/56 65/56
12/48 20/48 53/48
14/56 36/56 65/56
36/72 81/72 98/72

But I haven't found a way to generate them, other than trying.
I suspect there are probably infinitely many even of just the form 1/d, d/d, a/d; which might be easier to find a formula for. a/d seems to converge on 4 (with a notable outlier)

1/9 9/9 40/9
1/12 12/12 52/12
1/28 28/28 116/28
1/33 33/33 136/33
1/44 44/44 980/44
1/57 57/57 232/57
1/64 64/64 260/64
1/96 96/96 388/96
1/105 105/105 424/105
1/145 145/145 584/145
1/156 156/156 628/156
1/204 204/204 820/204
1/217 217/217 872/217
1/273 273/273 1096/273
1/288 288/288 1156/288
1/352 352/352 1412/352
1/369 369/369 1480/369
1/441 441/441 1768/441
1/460 460/460 1844/460
1/540 540/540 2164/540
1/561 561/561 2248/561
1/649 649/649 2600/649
1/672 672/672 2692/672
1/768 768/768 3076/768
1/793 793/793 3176/793
1/897 897/897 3592/897
1/924 924/924 3700/924
1/1036 1036/1036 4148/1036
1/1065 1065/1065 4264/1065
1/1185 1185/1185 4744/1185
1/1216 1216/1216 4868/1216
1/1344 1344/1344 5380/1344
1/1377 1377/1377 5512/1377
1/1513 1513/1513 6056/1513
1/1548 1548/1548 6196/1548
1/1692 1692/1692 6772/1692
1/1729 1729/1729 6920/1729
1/1881 1881/1881 7528/1881
1/1920 1920/1920 7684/1920
1/2080 2080/2080 8324/2080
1/2121 2121/2121 8488/2121
1/2289 2289/2289 9160/2289
1/2332 2332/2332 9332/2332
Title: Re: three square numbers
Post by towr on Aug 18th, 2013, 8:06am
One generic solution is 1/d, d/d, (4d+4)/d, with d=5k2+/-4k

1/d + 4(d+1)/d2 = [d + 4(d+1)]/d2 = [5d + 4]/d2 = [25k2+/-20k + 4]/d2 = [(5k +/- 2)/d]2
1 + 4(d+1)/d2 = [d2 + 4(d+1)]/d2 = [(d+2)/d]2
4(d+1)/d + 4(d+1)/d2 = [4(d+1)d + 4(d+1)]/d^2 = [2(d+1)/d]2


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