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How can I show that if $\text{gcd}(a, b, c) = 1$ then there exists $k$ such that $\text{gcd}(a, b+kc)=1$ ?

I tried to use Bezout's relation but couldn't get to keep a unit in front of $b$.

Context: I saw this used in a proof to show that $SL_2(\mathbb{Z}) \to SL_2(\mathbb{Z}/n\mathbb{Z})$ is surjective.

it's a hire car baby
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Thomas
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2 Answers2

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You can prove your claim as follows.

Suppose $\gcd(a,b)=1$, then the claim is obviously true, e.g. by putting $k=0$ (or $k=a$ if you only want positive $k$s).

So suppose $\gcd(a,b)=h>1$, then we can write $a=xh$ and $b=yh$ for some $x$ and $y$. Suppose for a contradiction that the claim is false, in particular suppose that $\gcd(a,b+xc)=z>1$. If $z|h$, then $z$ cannot divide $c$, otherwise $\gcd(a,b,c)\geq z>1$, a contradiction. Then $z|x$. Moreover, if $z|x$, then $z|b$, and so $z|h$. This means that the claim can be false only if $x$ and $h$ are not coprime.

Then, let us consider $\gcd(a,b+c\frac{x}{\gcd(x,h)})$, and suppose it is equal to $w>1$. Now, suppose that $w|h$: then $w$ cannot divide $\frac{x}{\gcd(x,h)}$, so $w|c$, but this contradicts $\gcd(a,b,c)=1$. Then $w|x$ and $w$ does not divide $h$, but then $w|b$, so $w|hy$, which implies that $w|y$, and this contradicts our definition of $h$, since $hw$ divides both $a$ and $b$.

Leo163
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0

If $gcd(a, b,c)=1$, then there is no common factor among the three except $1$.

Let

$$gcd(a,b+kc)\neq1$$ be equation $1$.

Now in the equation $gcd(a, b+kc)$, only two out of three can at max have a non unity $gcd$ as if two pairs had a non unity $gcd$, then the lower of the two $gcd$s would have been the $gcd$ of the trio. Meaning - only one of the pair $(a,b)$, $(b,c)$ and $(c,a)$ can have a non unity $gcd$.

So if $gcd(b,c)=r$, equation $1$ will hold as $a$ cannot be divisible by $r$ and the second term can be made divisible by $a$ by putting $k=a$ or it's multiples. This will result in second term having $gcd$ equal to $a$ with $a$ whereas the first term, $b$ has a $gcd$ of $1$ with $a$ resulting in a net $gcd$ of $1$.

If $gcd(a,b)=r$, equation $1$ will hold as $c$ is not divisible by $r$ and you can simply put $k=nr+1$ where $n=0,1,2...$

If $gcd(a,c)=r$, equation $1$ will hold as $b$ is not divisible by $r$ and you can put $k$ as $0$ to confirm.

EDIT - I did not phrase the second paragraph properly. I meant that any $gcd(x_1,gcd(x_2,x_3))$ will always be $1$ where $x_1,x_2,x_3$ are distinct members of the set $S={a,b,c}$

Sauhard Sharma
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