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Prove gcd(x,y) = gcd(x-y,y) for x > y Here is my work so far:

If gcd(x,y) = d, then we can denote x = ad , y = bd, so x - y = (ad - bd) = d(a - b), so (x-y) is a multiple of d, so gcd(x -y, y) = gcd(d(a - b), bd) = d, so gcd(x-y, y) = gcd(x,y).

My teacher challenged it by asking to show why gcd(d(a - b), bd) = d.

I think that gcd(d(a - b), bd) = d if gcd(a-b, b) = 1 --> gcd(a,b) = 1,

But I need some more help as to why this occurs, or if their is a simpler method of proving gcd(x,y) = gcd(x-y,y) for x > y

Bill Dubuque
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Knocker379
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2 Answers2

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Suppose $z \mid x$ and $z \mid y$. Then write $x = az$, $y = bz$. Then $x - y = az - bz = (a - b) z$, so $z \mid x - y$. So every common factor of $x$ and $y$ is a common factor of $x - y$ and $y$.

Now suppose $z \mid x - y$ and $z \mid y$. Then write $x - y = az$, $y = bz$. Then $x = (x - y) + y = az + bz = (a + b) z$, so $z \mid x$. So every common factor of $x - y$ and $y$ is a common factor of $x$ and $y$.

Since $x, y$ and $x - y, y$ have the same set of common factors, they must have the same gcd.

Michael Hardy
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Mark Saving
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  • @Knocker379 I literally just proved that. It's the first paragraph. – Mark Saving Jul 19 '21 at 17:51
  • Thanks it took me a second to get that – Knocker379 Jul 19 '21 at 17:54
  • Notice the typographical difference between $x | y$ (coded as x | y) and $x\mid y$ (coded as x\mid y). Also notice what $x\nmid y$ looks like (coded as x\nmid y) as opposed to something like $x\not|y$ (coded as x\not|y). $$ \begin{align} & x\mid y \ {} \ & x\nmid y \ {} \ & x | y \ {} \ & x\not|y \end{align} $$ – Michael Hardy Jul 19 '21 at 17:58
  • Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jul 20 '21 at 01:35
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  • First prove that every divisor that $x$ and $y$ have in common is a divisor of $x-y.$
  • Then prove that every divisor that $x-y$ and $y$ have in common is a divisor of $x.$

Thus the set of common divisors of $x$ and $y$ is the same as the set of common divisors of $x-y$ and $y.$

Michael Hardy
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    Please strive not to add more dupe answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Jul 20 '21 at 01:35
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    @BillDubuque : I'm not aware of any duplication of anything by this answer. – Michael Hardy Jul 20 '21 at 05:35
  • Did you even look? It's a special case of this answer, and likely hundreds of others, as is the question. I don't know how any experienced user could conceive that this question (and answer) was not a dupe. – Bill Dubuque Jul 20 '21 at 17:09
  • @BillDubuque : I didn't look since you didn't provide a link. Until you did. – Michael Hardy Jul 20 '21 at 17:32
  • @BillDubuque : I regard it as an improvement over the answer to which you linked, since it's simpler. Making things simple takes some work and is worthwhile. – Michael Hardy Jul 20 '21 at 17:34
  • That answer is the accepted answer in the linked dupe on the qiuestion, so it's not clear why you did not look there. It is hardly "some work" to specialize $q=-1$. – Bill Dubuque Jul 20 '21 at 17:39
  • @BillDubuque : There wasn't any linked duplicate at the time I posted this answer. – Michael Hardy Jul 20 '21 at 17:43
  • Please follow the link in my first comment to the meta post. Answering obvious dupes like this will likely result in suspension (and already has for many users who continue to do so after being notified about such). – Bill Dubuque Jul 20 '21 at 17:48