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 Add To Cube, Get Square? (Posted on 2007-08-03)
Is there any perfect cube that becomes a perfect square if you add 7 to it?

 See The Solution Submitted by K Sengupta Rating: 4.0000 (1 votes)

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 re: thoughts | Comment 2 of 4 |
(In reply to thoughts by Charlie)

Also, checking the numbers up to 10,000,000, 46 is the highest whose cube can be incremented by a single-digit number to get a perfect square:

1 ^ 3 + 3 = 2 ^ 2
2 ^ 3 + 1 = 3 ^ 2
3 ^ 3 + 9 = 6 ^ 2
6 ^ 3 + 9 = 15 ^ 2
40 ^ 3 + 9 = 253 ^ 2
46 ^ 3 + 8 = 312 ^ 2

This might lead to a clue as to the proof needed: that the smallest difference (other than zero) possible might increase as the numbers being cubed get larger.

If we allow the increment to be up to 100, the possibilities seem to be cut off at 8158^3:

1 ^ 3 + 3 = 2 ^ 2
2 ^ 3 + 1 = 3 ^ 2
3 ^ 3 + 9 = 6 ^ 2
4 ^ 3 + 17 = 9 ^ 2
5 ^ 3 + 19 = 12 ^ 2
6 ^ 3 + 9 = 15 ^ 2
7 ^ 3 + 18 = 19 ^ 2
8 ^ 3 + 17 = 23 ^ 2
9 ^ 3 + 55 = 28 ^ 2
10 ^ 3 + 24 = 32 ^ 2
11 ^ 3 + 38 = 37 ^ 2
12 ^ 3 + 36 = 42 ^ 2
13 ^ 3 + 12 = 47 ^ 2
14 ^ 3 + 65 = 53 ^ 2
18 ^ 3 + 97 = 77 ^ 2
19 ^ 3 + 30 = 83 ^ 2
20 ^ 3 + 100 = 90 ^ 2
24 ^ 3 + 100 = 118 ^ 2
40 ^ 3 + 9 = 253 ^ 2
43 ^ 3 + 17 = 282 ^ 2
44 ^ 3 + 80 = 292 ^ 2
45 ^ 3 + 79 = 302 ^ 2
46 ^ 3 + 8 = 312 ^ 2
52 ^ 3 + 17 = 375 ^ 2
55 ^ 3 + 89 = 408 ^ 2
72 ^ 3 + 73 = 611 ^ 2
109 ^ 3 + 15 = 1138 ^ 2
152 ^ 3 + 68 = 1874 ^ 2
243 ^ 3 + 37 = 3788 ^ 2
356 ^ 3 + 73 = 6717 ^ 2
584 ^ 3 + 65 = 14113 ^ 2
2660 ^ 3 + 100 = 137190 ^ 2
5234 ^ 3 + 17 = 378661 ^ 2
8158 ^ 3 + 24 = 736844 ^ 2

Edited on August 3, 2007, 10:19 am
 Posted by Charlie on 2007-08-03 10:07:33

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