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January 7th, 2015, 03:06 AM   #1
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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∈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?
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January 7th, 2015, 07:30 AM   #2
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Looks good to me.
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odd, solve, x38y5

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