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Three Term Trial (Posted on 2011-01-15) Difficulty: 3 of 5
Prove that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic sequence.

No Solution Yet Submitted by K Sengupta    
Rating: 2.5000 (2 votes)

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Some Thoughts Restating the Problem | Comment 1 of 4
Well obviously, the cube roots of primes are not integers and are not even rational.  But that doesn't mean that these three irrational numbers can't form an arithmetic sequence.  Any three real numbers are terms (not necessarily consecutive) of an infinite number of arithmetic sequences, as long as the ratio of their differences are rational.

So, what we are being asked to prove is that there do not exist three prime cube roots, x, y, and z, such that (x-y)/(x-z) is rational.

Or in other words, there do not exist integers P and Q such that 
P(x-y) = Q(x-z), where x, y and z are cube roots of distinct primes.

Unfortunately, I have no idea how to proceed from here.

  Posted by Steve Herman on 2011-01-15 23:38:03
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