Prime factors of $x$ and $y$ assuming ${\rm lcm}(x,y) = \gcd(x,y)^2$

Question

I know there is a relation between the $\gcd (x, y)$ and $\operatorname{lcm} (x, y)$ like this: $$ \operatorname{lcm} (x, y) \gcd (x, y) = xy. $$ E.g: $x = 588$ and $y = 616$

$588 = 2 ^ 2 \cdot 3 \cdot 7 ^ 2$

$616 = 2 ^ 3 \cdot 7 \cdot 11$

$\gcd (x, y) = 2 ^ 2 \cdot 7 = 28$

$\operatorname{lcm} (x, y) = 2 ^ 3 \cdot 3 \cdot 7 ^ 2 \cdot 11 = 12936$

In this case $\gcd (x, y) ^ 2$ is not equal to $\operatorname{lcm} (x, y)$

So how can I find the prime factors of $x$ and $y$ if $\operatorname{lcm} (x, y) = \gcd (x, y) ^ 2$?

Answer 1

Let $p$ be a prime dividing $x$. If $p$ doesn't divide $y$, then $p$ divides $\DeclareMathOperator{\lcm}{lcm}\lcm(x,y)$, but not $\gcd(x,y)$, which is a contradiction. Similarly, each prime divisor of $y$ also divides $x$.

Hence $x=p_1^{m_1}p_2^{m_2}\dotsm p_k^{m_k}$ and $y=p_1^{n_1}p_2^{n_2}\dotsm p_k^{n_k}$, where $m_i\ge1$ and $n_i\ge 1$, with $p_1,\dots,p_k$ distinct primes ($k$ could be $0$, in which case $x=y=1$).

Note that \begin{align} \gcd(x,y)&=p_1^{\min(m_1,n_1)}p_2^{\min(m_2,n_2)}\dotsm p_k^{\min(m_k,n_k)} \\[4px] \lcm(x,y)&=p_1^{\max(m_1,n_1)}p_2^{\max(m_2,n_2)}\dotsm p_k^{\max(m_k,n_k)} \end{align} so for each $i$ we must have $$ \max(m_i,n_i)=2\min(m_i,n_i) $$ and that's all.

You get all numbers with this property with the following recipe. Take $p_1,\dots,p_r,q_1,\dots,q_s$ arbitrary pairwise distinct primes (it can even be $r=0$ or $s=0$; an empty product is, by definition, $1$). Take $a_1,\dots,a_r,b_1,\dots,b_s$ arbitrary integers $\ge1$. If $$ A=p_1^{a_1}\dotsm p_r^{a_r} \qquad B=q_1^{b_1}\dotsm q_s^{b_s} $$ then $x=A^2B$ and $y=AB^2$ are numbers such that $\lcm(x,y)=\gcd(x,y)^2$. Note also that any two numbers $x$ and $y$ with this property are of this form.

Answer 2

We have that: $$\text{lcm}(x,y)\text{gcd}(x,y) = xy$$ So: $$\text{lcm}(x,y) = \text{gcd}(x,y)^2\implies\text{lcm}(x,y)\text{gcd}(x,y) = \text{gcd}(x,y)^3$$ The LHS of this is just $xy$, so we have that: $$xy = \text{gcd}(x,y)^3$$ Now, consider some specific prime $p$. If this occurs in either $x$ or $y$, we have that it must occur in both (otherwise LHS $\mod p\equiv 0$, but RHS $\mod p\not\equiv 0$).

Say that $p^k$ is the highest power of $p$ that occurs in both $x$ and $y$. Then, WLOG let $p^k\mid x$, and $p^{k+j}\mid y$ be the highest powers of $p$ in each productand. Then, we have that: $$x^{2k+j} \equiv x^{3k}$$ So, $j = k$, and $p^{2k}\mid y$ is the highest power of $p$ dividing $y$.

So:

1. Factor $\text{gcd}(x,y)$ into $p_1^{e_1}\dots p_k^{e_k}$
2. For each $i$, find if $p_i^{2e_i}$ divides $x$ or $y$ (it'll have to divide one). Then $p_i^{e_i}$ divides the other.
3. Repeat for all $i$.

It's possible there are other relations you can pull from this condition, but this is all I see.