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My solution is as follows:

Let $d_1= (a,b) > 0$. Then by Bezout's identity, there exists integers $x_1$and $y_1$ such that, $d_1=ax_1+by_1$. Similarly let $d_2 = (a+c,b) > 0$ then there exists integers $x_2$ and $y_2$ such that, $d_2=(a+c)x_2+by_2$. Then since $b\mid c$ there exists integer $k$ such that $c=bk$ for some $k$.

Note that, $d_2=(a+c)x_2+by_2$. Since $c = bk$ then, $d_2=(a+bk)x_2+by_2$ $\Rightarrow$ $d_2 = ax_2+b(kx_2+y_2)$, which is a linear combination in $a$ and $b$. Since $d_1 = (a,b)$, it divides any linear combination in $a$ and $b$. Thus $d_1\mid d_2$.

Similarly, $d_1=ax_1+by_1$ $\Rightarrow$ $d_1=ax_1+by_1+cx_1-cx_1$. Since, $c=bk$ $\Rightarrow$ $d_1=(a+c)x_1+b(y_1-kx_1)$. Which is a linear combination in $a+c$ and $b$. Thus $d_2\mid d_1$.

Since $d_1$ and $d_2$ are $\gcd$ of two integers, and $d_1\mid d_2$ and $d_2\mid d_1$, we conclude that $d_1=d_2$ as desired.

Thank you for reading upto here. If there is any errors feel free to comment.

Kritesh Dhakal
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  • You need to use $,d_1,d_2>0$ to conclude $,d_1 = d_2,$ (cf. gcd unit normalization). Else it is fine, but it is overkill to use Bezout, e.g. see my proof in the dupe, which uses only divsiibility properties so works in domains where Bezout fails, e.g. polynomial rings like $,\Bbb Z[x],\ \Bbb Q[x,y].\ $ – Bill Dubuque Aug 23 '20 at 21:08

1 Answers1

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Seems fine.

Notice that we have

$$\gcd(a+b,b)=\gcd(a, b)$$

WLOG, if $c=kb$, $k \ge 1, k \in \mathbb{Z}$, then by induction, we have

\begin{align}\gcd(a+c,b)&=\gcd(a+kb, b)\\&= \gcd(a+kb-b, b)\\&=\gcd(a+(k-1)b, b)\\ &= \gcd(a,b) \end{align}

Siong Thye Goh
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