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I am not really sure if it is correct. My idea is:

Let $ab=x^3 $ and $(a, b) = 1$ and $a, b, x$ are some elements in an arbitrary ring $R$.

Then $a|x^3 $ and $b|x^3$ which means there exists some elements $c,d \in R$ that satisfies $ac = x^3$ and $bd = x^3$. We then know that $a|d$ because $a, b$ are coprime and $a\space|\space x^3 = bd$.

For example: Consider the ring $R=Z[i]$

I want to find all solutions $x,y \in \mathbb{Z}$ with $y^2=x^3-1$

Now the factored form in $R$ is $$x^3=(y+i)(y-i) $$

I would like to know why the factors have to be cubes .

İbrahim İpek
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AnabolicHorse
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    I think you mean "the product of two relatively prime positive integers is a cube." Because if product is prime it is not a cube. – coffeemath Jul 01 '19 at 19:38
  • This OP says "in a ring" ... so does it work that generally? The guy who claims this is a duplicate: Do you claim one of the answers in that other question works in a general ring? – GEdgar Jul 01 '19 at 19:43
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    You are on the right track for a ring that happens to enjoy unique factorization but not in general. The concept of greatest common divisor does not make sense in an arbitrary ring. – Ethan Bolker Jul 01 '19 at 19:44
  • Even if "greatest common divisor" doesn't work in a ring, perhaps "relatively prime" does. – GEdgar Jul 01 '19 at 19:46
  • @GEdgar Perhaps, modulo a definition of "relatively prime" in the ring under consideration. My point is that you must specify definitions and assumptions before the question even makes sense. – Ethan Bolker Jul 01 '19 at 19:56
  • It's not true in arbitrary rings. The proof in the dupe is probably the simplest general result. If you are looking for something else then please explain more precisely what class of rings you are interested in and we can reopen it. – Bill Dubuque Jul 01 '19 at 20:11
  • Thanks for helping . The proof in the dupe was exactly what I was lookung for . – AnabolicHorse Jul 01 '19 at 20:14
  • No more answers allowed. This fails in the ring $\mathbb Z[\sqrt{-5}]$. – GEdgar Jul 01 '19 at 20:18
  • @GEdgar Counterexamples are easy since it is generally false in non UFD/GCDs, e.g. a generic counterexample $,xy = z^3,$ in $,\Bbb Z[x,y,z]/(xy-z^3)\ \ $ – Bill Dubuque Jul 01 '19 at 22:19

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If the $a, b$ are relatively prime, it means they contain no common prime factors. Since we know that,

$$ab = x^3$$

We also know that all exponents of the primes $ab$ contain, must be a multiple of $3$. Because $a$ and $b$ have no common prime factor, their all prime factors have to be exponentiated to a multiple of $3$. So they are cubes. QED.

İbrahim İpek
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