Exercise 1.2.26 (Solvability of $x + y = s, \ x \wedge y = g$)

Question

Let $s$ and $g>0$ be given integers. Prove that integers $x$ and $y$ exist satisfying $x+y = s$ and $(x,y) = g$ if and on y if $g \mid s$.

Richard Ganaye · Accepted Answer

\begin{proof} 
\begin{itemize}
\item If integers $x$ and $y$ exist satisfying $x+y = s$ and $(x,y) = g$, then $g \mid x, g \mid y$, thus $g \mid x + y$, so $g \mid s$.

\item Conversely, assume that $g \mid s$. Then (Theorem 1.9 and Problem 18),
$$g \wedge (s - g) = g \wedge s = g.$$
Therefore $x = s-g$ and $y = g$ satisfy $x + y = s, \ x \wedge y = g$.
\end{itemize}
\end{proof}

Exercise 1.2.26 (Solvability of $x + y = s, \ x \wedge y = g$)

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Exercise 1.2.26 (Solvability of $x + y = s, \ x \wedge y = g$)

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