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Problem: Given integers $a$ and $b$ prove that if there exist integers $x$ and $y$ for which $ax+by=(a,b)$ then $(x,y)=1.$ Note that $(a,b)=\gcd(a,b).$

Proof: Suppose $(x,y)=m,$ where $m\in\mathbb{Z}.$ Then $y=mk_1$ and $x=mk_2$ for some $k_1, k_2 \in \mathbb{Z}$. Let $(a,b)=d$. We now have $m((a/d)k_1+(b/d)k_2)=1.$ Since both numbers are integers we can conclude that $m=1$ which completes the proof.

Remark: I am unsure about the last claim and would therefore, be grateful if someone could either rectify my proof or declare that it is correct.

Clyde Kertzer
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1 Answers1

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Your proof is correct. Here is an alternative. First of all, note that if $ax+by=c$ (where $a$, $b$, $c$, $x$ and $y$ are integers), then $(x,y)$ divides $c$ (since it has to divide any integer linear combination of $x$ and $y$).

Assume now that $(a,b)=1$ and $ax+by=1$. Then $(x,y)$ divides $1$, hence $(x,y)=1$. Now, if $d=(a,b)\neq 1$, then $\frac{a}{d}x+\frac{b}{d}y=1$ and $\frac{a}{d}$ and $\frac{b}{d}$ are integers, so $(x,y)=1$ by the previous case.

Taladris
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