Question

Suppose gcd(2p, p^2 + 3q^2)=1. Since 2p(p^2 +3q^2) is a cube and gcd(2p, p^2 + 3q^2) = 1, by "what" there exists an integer "u" so that u^3 = p^2 + 3q^2?

Answer 1

my attempt at this goes something like this:
you are told that this is a cube:\[w^3=2p(p^2+3q^2)\]so, for this to also be a cube:\[u^3=p^2+3q^2\]2p must be equal to 1 (so p=1/2) as then w^3 would be of the form:\[w^3=2p(p^2+3q^2)=2*\frac{1}{2}((\frac{1}{2})^2+3q^2)=\frac{1}{4}+3q^2=\frac{1+12q^2}{4}\]for this to be an integer:\[1+12q^2=4n\]so:\[q^2=\frac{4n-1}{12}\]so, in conclusion, for u^3 to be an integer:\[p=\frac{1}{2}\]and\[q^2=\frac{4n-1}{12}\]

Answer 2

what method are you using? like what theorem?

Answer 3

no particular theorem, just trying to use logical thinking.

Answer 4

i need to know what theorem for this questions, not nessarily logic

Answer 5

sorry - can't help you there - I just enjoy maths as a hobby and have no formal training on it beyond GCSE A-Levels (age 18 maths exams in UK)