Prove that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic sequence.
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 20110116 06:07:44 