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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 possible solution | Comment 3 of 4 |

The question requires that the cube root of some prime, P, be of a form which is partly rational(x/y) and partly irrational(I) i.e. I*x/y,(cancelling until x and y are relatively prime) so that the three numbers (I^3*x^3)/y^3, (I^3*m^3x^3)/y^3 and (I^3*n^3x^3)/y^3 are distinct primes.

Even to produce the first term, we need I^3 to be a rational number, say k/l (cancelling until k and l are relatively prime) so that k/l*x^3/y^3=P, but then:
1. kx^3= Ply^3.
2. Assume that y>x: x^3= (Ply^3)/k. Then k divides P, but P is prime; or k divides l, but k and l are relatively prime; or k divides y^3, but x and y are relatively prime.
3. Assume that x>y: kx^3/Pl=y^3. By parity of reasoning, l cannot divide either k or x, and P cannot divide x. So P must divide k, say, t times; k=Pt
4. But then tx^3=ly^3, and again by parity of reasoning, t, a factor of k, cannot be a factor of y or l. So there is no way to produce the first term that starts the series.

  Posted by broll on 2011-01-16 06:07:44
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