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I was solving a math problem and the book at some point wrote that

$$\gcd(44m+15;15m+5) = \gcd(m;5)$$ when $m$ is an integer.

Why?

Thomas Andrews
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TechnoKnight
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    $$-(44m+15)+3(15m+5)=m\ 15(44m+15)-44(15m+5)=5.$$ – Thomas Andrews Jan 13 '22 at 01:47
  • @ThomasAndrews I might be stupid but isn't this incorrect? The property says $gcd(n,m)=gcd(n−km,m)$ right? However, as an example, here it's no longer $n-km$ since you multiplied each side by a number not only one. so it's more like $an-km$ right now – TechnoKnight Jan 13 '22 at 02:09
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    That wasn't what I was hinting at with my comment. If $ax+by=d,$ then $\gcd(a,b)\mid d.$ – Thomas Andrews Jan 13 '22 at 02:24
  • @ThomasAndrews But the property I wrote works just as fine, right? – TechnoKnight Jan 13 '22 at 02:38
  • Use the reduction step of the Euclidean algorithm to continually reduce the coef's of $m$, e.g. see here in the first dupe for a similar gcd calculation. – Bill Dubuque Jan 13 '22 at 03:24
  • Or $,(\color{#c00}{44}m+\color{#0a0}{15},,\color{#0a0}3m+\color{#c00}1)=1,$ by $,\color{#0a0}{15(3)}-\color{#c00}{44(1)}=1,$ and the formula here, so Euclid's Lemma $,\Rightarrow (44m+15,,5(3m+1) = (\color{#90f}{44m+15},5) = (\color{#90f}{-m},5) = (m,5),$ by $,\color{#90f}{(44m+15)\bmod\color{#000}{5} = -m},$ and gcd modular reduction, i.e. the descent step of the Euclidean algorithm. $\ \ $ – Bill Dubuque Jan 13 '22 at 05:14

1 Answers1

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$$\gcd(44m+15, 15m+5)=\gcd(-m,15m+5)=\gcd(-m, 5)=\gcd(m,5)$$

The first equalities derive from the property of the gcd that states: $\gcd(n,m)=\gcd(n-km, m)$, where $k$ is an integer. The last equality is trivial since the divisors of $m$ are the same as the divisors of $-m$.

MH.Lee
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davinci_07
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