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I want to prove $((a,b),b)=(a,b)$ for $a$, $b$ integers with some formal mathematical statement like Bezout's identity i.e., $(a,b)=d=am+bn$ or something else.

Any ideas?

Here $(a,b)$ means gcd.

My first sight is let $(a,b)=d$ and factors $a=d q_a$, $b=d q_b$ and proceed.

I want more rigorous proof

Martin Sleziak
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phy_math
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4 Answers4

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Hint $\,\ ((a,b),b) = (a,(b,b)) = (a,b)\ $ by the gcd Associative Law.

Or directly $\,\ d\mid((a,b),b)\iff d\mid(a,b),b\iff d\mid a,b,\, b\iff d\mid a,b\iff d\mid (a,b)\,$ where we used the gcd Universal Property $3$ times.

Or $\ d\mid b\,\Rightarrow\, (d,b) = (d, b\bmod d) = (d,0) = d\,$ by the Euclidean algorithm, or $\,(d,b) = d(1,b/d) = d\,$ by the gcd Distributive Law.

Or use $\ (a\Bbb Z + b\Bbb Z) + b\Bbb Z = a\Bbb Z + (b\Bbb Z + b\Bbb Z) = a\Bbb Z + b\Bbb Z\ $ and Bezout.

Bill Dubuque
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Just note that $d=(a,b)$ is a divisor of $b$ and that $(x,y)=x$ whenever $x\mid y$.

timon92
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Hint for proof with Bezout: because $\gcd(a,b)$ is a linear combination of $a$ and $b$, any linear combination of $\gcd(a,b)$ and $b$ is also a linear combination of $a$ and $b$.

Arthur
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  • You need to say much more than that to obtain a valid proof that way. – Bill Dubuque Sep 21 '17 at 13:59
  • That would make this into a full answer, and not just a hint. I called it a hint, and I stand by it. – Arthur Sep 21 '17 at 14:00
  • So precisely what is your claim, that any linear combination of $a$ and $b$ is equal to the gcd? We can't read your mind, nor should we have to. – Bill Dubuque Sep 21 '17 at 14:07
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    @BillDubuque No, my claim is exactly what I say: any number which is a linear combination of $\gcd(a,b)$ and $b$ may be realised as a linear combination of $a$ and $b$. – Arthur Sep 21 '17 at 14:59
  • But you haven't explained at all how that has any relation to the problem at hand. We should have higher standards for answers. Otherwise users can post as an "answer" motley wild guesses that lead nowhere (and, alas, that does occur not too infrequently). – Bill Dubuque Sep 21 '17 at 16:13
  • @BillDubuque I don't know what more you want. The OP wanted to do it using Bezout's, and I gave a pointer to how that can be done. Bezout's lemma is all about the connection between possible linear combinations and the $\gcd$. I say explicitly that it is about Bezout's, and I point out and argue for an explicit connection between the linear combinations of one pair of numbers and the linear combinations of another pair of numbers. The exact two pairs, mind you, that we're asked about. I specifically didn't want to write an answer that only left a few cracks for the OP to fill in. – Arthur Sep 21 '17 at 21:44
  • What you deem to be "a few cracks", I deem "huge gaps". I wonder if you really know how to complete the proof. You may well be missing something that is revealed when you attempt to proceed from your first step. I can't say for sure since I have no idea how you plan to proceed. – Bill Dubuque Sep 22 '17 at 01:01
  • @BillDubuque My last sentence was "I didn't want to leave just a few cracks", but you seem to have missed the negative in there. Also, since you genuinely seem to not be able to grasp the approach, here is a "hint" of your type, borrowing group theory notation purely for convenience: if $d=\gcd(a,b)$ then $$\langle a,b\rangle=\langle d\rangle\subseteq \langle d,b \rangle\subseteq\langle a,b\rangle$$ Bezout's is exactly the first equality, my hint is the last containment. – Arthur Sep 22 '17 at 04:49
  • I don't like to call that a hint, because I think hints should leave something more than routine calculations. A hint should only point you in the right direction, and there should still be some thinking left to do once you've read the hint. If there isn't, then what you have is more substantial than a hint in my opinion. – Arthur Sep 22 '17 at 04:55
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$$ \exists x, y \in \mathbb{Z} : (a, b)x+by= ((a, b), b)$$

$$(a, b) \mid (a, b) \land (a, b) \mid b \Longrightarrow (a, b) \mid (a, b)x+by=((a, b), b)$$

$$(a, b) \mid ((a, b), b) \land ((a, b), b) \mid (a, b) \Longrightarrow (a, b) = ((a, b), b)$$

For the last step you need that both $((a, b), b)$ and $(a, b)$ are positive because otherwise you could get that $((a, b), b)= -(a, b)$.

Sofía Marlasca Aparicio
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