January 7th, 2015, 03:06 AM  #1 
Member Joined: Oct 2013 Posts: 36 Thanks: 0  solve x^3=8y^5 for odd y
"Find all pairs of positive integers x, y ∈ N such that x^3 = 8y^5 and y is odd. [You may quote the fundamental theorem of arithmetic without proof.]" this was the question and here is what i did: x=2y^(5/3) => x∈N only if y^(1/3) ∈N =>claim: y is a perfect cube proof: suppose y is not a perfect cube ie, y=/=n^3 for n ∈N, and set y^(1/3)=a/b for a,b∈N and a/b is in lowest form. => y=a^3/b^3 by the fundamental theorem of arithmetic the unique prime factorisation of a exists and so the prime factorisation of a^3 is equal to the prime factorisation of a but with each power tripled. since b cannot be 1 (since we assumed y was not a perfect cube) any prime which divides b and therefore b^3 cannot divide a^3 (lowest form a/b) => a^3/b^3=/=an integer. so if y is an integer then y^(1/3) is rational (and also clearly natural) only if y is a perfect cube. => since y is odd y=(2k+1)^3 for k∈N => x=2(2k+1)^5. so these are all the solutions. is this an acceptable answer to the question? 
January 7th, 2015, 07:30 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
Looks good to me.


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odd, solve, x38y5 
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