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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)

Comments: ( Back to comment list | You must be logged in to post comments.)
re: Restating the Problem | Comment 2 of 4 |
(In reply to Restating the Problem by Steve Herman)

proof can be completed by seeing that if you take
x=p^(1/3) y=q^(1/3) z=r^(1/3) and n some rational number
with p,q,r prime numbers
then we need
p^(1/3) - q^(1/3)
--------------------- = n
p^(1/3) - r^(1/3)

this can be simplified by multiplying the left side by
p^(2/3) + (pr)^(1/3) + r^(2/3)
-------------------------------------
p^(2/3) + (pr)^(1/3) + r^(2/3)

now you are left with a integer in the denominator but with
the sum of several cube roots in the numerator.  Each of these cube roots are irrational and thus their sum is irrational and thus
(x-y)/(x-z) can not be rational

QED

  Posted by Daniel on 2011-01-16 05:35:50

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