Solve $x+y=84, \; \operatorname{lcm}(x,y) = \big( \gcd(x,y) \big)^2$

Question

I am interested in solving the system of equations

$$ (S) \; \left\{ \begin{array}{l} x + y = 84 \\ \operatorname{lcm}(x,y) = \big( \gcd(x,y) \big)^2 \end{array} \right. $$

where $(x,y) \in \mathbb{N}^{\ast}$.

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If $d = \gcd(x,y)$, then $d \mid 84$. Therefore:

$$ d \in \lbrace 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84 \rbrace. $$

Therefore, for each possible value $d$, we have to solve the system:

$$ (S) \; \left\{ \begin{array}{l} x + y = 84 \\ \operatorname{lcm}(x,y) = d^2 \end{array} \right. $$

However, I'm not sure how to solve such a system given $x+y$ and $\operatorname{lcm}(x,y)$.

Answer 1

As @Alexander pointed out in the comments, $\operatorname{lcm}(x, y) = d^2 \iff xy = \operatorname{lcm}(x, y) \operatorname{gcd}(x, y) = d^2d = d^3$, so $ y = \frac{d^3} x \implies x^2-84x+d^3=0 $ and it's easy to carry on from here.

Hope it helps.