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I studied that $$\gcd(a,b) = \gcd(a, a+bk)$$ where $k$ belongs to an integer.

Now suppose $a = 8$, $b = 15$, and $k = 4$. So, $\gcd(8,15) = 1$, but $\gcd(8, 8+15(4)) = 4$, which doesn't satisfy this property.

Blue
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Navdeep Singh
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  • you can easily prove JJ Hoo's statement by taking $gcd(a,b)=p\Rightarrow a=px_1, b=px_2$ where $gcd(x_1,x_2)=1$ and $p,x_1,x_2 \in \Bbb{Z}$ – An Alien Oct 17 '21 at 07:15
  • There is a typo in the statement - see the linked dupe for the correct statement and proofs. – Bill Dubuque Oct 18 '21 at 15:27

1 Answers1

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The reason that you have found a counterexample is because the property you stated is incorrect. I believe you mean the following property instead, for $k\in\mathbb{Z}$: $$gcd(a,b)=gcd(a,b+ak)$$ A proof follows from definitions you have likely learned, and I encourage you to attempt such a proof to ensure that you understand why this must be true!

JJ Hoo
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