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Progressively Geometric (Posted on 2009-09-19) Difficulty: 2 of 5
Prove that 11, 12 and 13 can never be three terms (not necessarily consecutive) of a geometric progression, irrespective of whether the common ratio is real or complex.

No Solution Yet Submitted by K Sengupta    
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Some Thoughts Flaw in the proposed solution | Comment 2 of 4 |
Daniel:

Before I find fault with your proposed proof, let me say that I really appreciate your contributions to this site.  I hope that you keep contributing for a long time.  I haven't seen anybody with your ability to solve a problem using Mathematica programming.

Problem with the proof
----------------------
Your proposed proof is headed in the right direction, but it is not quite there.  It could be used incorrectly to prove that 3, 6, and 12 cannot be in the same geometric progression, because ln(6/3)/ln(12) is irrational.  Clearly, 3, 6 and 12 are in geometric progression.  The problem is that what you have really proved is that 1, 11, 12 and 13 cannot be in the same geometric progression, just as 1, 3, 6, 12 cannot be in the same geometric progression.

The problem is at the very start of the proof.  You cannot assume that 11 = r^x where r is the geometric ratio and x is integral.
You need to start as follows:
  11 = ar^x
  12 = ar^y
  13 = ar^z
Once you do, the proof doesn't work quite the way that you have done it.
I'll post the corrected proof separately.

Edited on September 20, 2009, 2:41 am
  Posted by Steve Herman on 2009-09-20 02:17:57

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