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I am doing my thesis on elliptic curves right now and in the meantime this lemma showed up:

Suppose $a$ and $b$ are integers such that $ab = m^3$ for some integer $m$. Let $d = \operatorname{gcd}(a,b)$, then we can write \begin{equation}\label{} a = d\cdot p_1^{r_1}\cdots p_t^{r_t}\cdot (\operatorname{integer})^3, \end{equation} where $p_i|d$ and $r_i \in \mathbb{Z}$.

So I tried taking a prime that does not divide $d$ and tried to show it must appear as a third power, however I couldn't figure it out. Any help would be appreciated!

Bill Dubuque
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Math4Life
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    Hint; For any prime $p$ the order to which $p$ divides $ab$ must be divisible by $3$. – lulu Apr 23 '20 at 16:58
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    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Apr 23 '20 at 17:15
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    @shaun Thank you! I will do that. – Math4Life Apr 23 '20 at 17:16
  • Please write an informative and specific question that describes your problem, not just the discipline that is relevant. – David G. Stork Apr 23 '20 at 17:37

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As lulu noted, the order of any prime dividing $ab$ must be divisible by 3, because $ab=m^3$. Now the product above just results from the fact that we can break up the primes dividing $m$ into those which divide only $a$, those which divide both $a$ and $b$, and thus divide $d$, and those which divide only $b$. Only the first two categories contribute to our expression for $a$, obviously, then you need simply observe that if a prime occurs exclusively in $a$, then the order of the prime in $a$ must be the same as the order of the prime dividing $ab$, whence it must be divisible by 3.

BelowAverageIntelligence
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Because for each prime $p$ that does not divide $b$ we have $v_p(a)= v_p(ab) = v_p(m^3)=3v_p(m)$.

We can clearly make the part that is not $dp_1^{r_1}\dots p_r^{r_t}$ equal to the product of all of the $p^{v_p(ab)}$ where $p$ is a prime not dividing $b$.

Asinomás
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Hint $ $ Let $\, c_a := d p_1^{r_1}\cdots p_t^{r_t}$ be the product of all prime factors of $\,a\,$ which also divide $\,b,\,$ and similarly for $\,c_b\,$ and $\,b.\,$ Then $\,a = a' c_a,\ b = c' c_b,\,$ and $\,a',b', c_a c_b\,$ are pairwise coprime, therefore $\, m^3 = ab = a'\, b'\, (c_a c_b)\,\Rightarrow\, a',b'$ are cubes too.

Bill Dubuque
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  • If you allow negative integers then you need to handle signs - which is easy. – Bill Dubuque Apr 23 '20 at 17:34
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    How do you know you can write $a$ like that? – Math4Life Apr 23 '20 at 17:35
  • @Math4Life I edited to clarify that part (I thought you were stuck on the other part based on what you wrote). Btw, we apply the linked Theorem twice, e.g. $,a(bc),$ a cube so $,a,bc,$ cubes, so $,b,c,$ cubes (i.e. the Theorem inductively extends to any number of pair coprime factors - it's a basic result often applied). – Bill Dubuque Apr 23 '20 at 18:19