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 Number Theory Number Theory Math Forum

 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. Tags odd, solve, x38y5 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post gen_shao Algebra 12 November 2nd, 2014 06:11 AM

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