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Subtract The Cubics, Get Product (Posted on 2007-07-22) Difficulty: 3 of 5
Determine all possible pairs of positive whole numbers (v, w) satisfying the following equation:

v3 – w3 = vw + 61

  Submitted by K Sengupta    
Rating: 3.3333 (3 votes)
Solution: (Hide)
v^3 – w^3 = vw + 61 > 0, giving:
v> w> = 1

Now, v^3 – w^3 = vw + 61
or, (v-w)(v^2 + vw + w^2) = vw + 61
or, (v^2 + w^2)*P + (P-1)vw = 61, where P = v-w
Now for P> = 3, we observe that:
v> = 3+w, so that:
3(v^2 + w^2) < = P(v^2+w^2)< 61
Or, v^2 + w^2) < = 20.
Since v> = w+3, this is only possible iff (v, w) = (4,1), but 4^3 – 1^3 = 63, while 4*3 + 61 = 73. This is a contradiction.

Hence P =1, 2

If P = 2, then 5*w^2 + 10*w – 53 = 0, which does not yield any integer root for w.

If P =1, then w^2+w – 30 = 0, so that;
w=5, ignoring the negative root which is inadmissible.

This yields v=6, and consequently (v, w) = (6,5) is the only possible solution to the given problem.

.................Q E D .....................

*** Also refer to an alternative methodology given by Brian Smith in this location .

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Another solutionBrian Smith2023-04-15 17:19:31
SolutionBrian Smith2007-07-23 10:35:35
No Subjectxdog2007-07-23 08:18:04
re: solutionFederico Kereki2007-07-23 00:19:46
solutionxdog2007-07-22 20:36:32
(Very) partial spoiler!Josie Faulkner2007-07-22 18:56:57
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