Exercise 1.2.30 (Solvability of $(x,y) = g,\ xy = b$)

Question

Let $b$ and $g>0$  be given integers. Prove that the equations $(x,y) = g$ and $xy = b$ can be solved simultaneously if and only if $g^2 \mid b$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $b$ and $g>0$  be given integers.
\begin{itemize}

\item
If $x \wedge y = g$ and $xy = b$, then $g \mid x$ and $g \mid y$, thus $x = \lambda g,\ y = \mu g$ for some integers $\lambda, \mu$. Then $b = xy = \lambda \mu g^2$, so $g^2 \mid b$.

\item Conversely, if $g^2 \mid b$, take $x = g,\ y = b/g$. Since $g\mid g^2$, a fortiori $g \mid b$, thus $x, y$ are integers, and $xy = g(b/g) = b$. Moreover, $g^2 \mid b$, thus $g \mid (b/g)$. By problem 18, using $g \mid (b/g)$,
$$x \wedge y = g \wedge \left( \frac{b}{g}\right) = g.$$
So  $x \wedge y = g, \ xy = b$ has at least the solution $x = g,\ y = b/g$.
\end{itemize}
\end{proof}

Exercise 1.2.30 (Solvability of $(x,y) = g,\ xy = b$)

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Exercise 1.2.30 (Solvability of $(x,y) = g,\ xy = b$)

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