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If $\gcd(a,b) =1$. Why does $\gcd(a+b,ab)=1$ ?

I know that if $\gcd(a,b)=1$ then there exists $u$ and $v$ where $au+bv=1$. But I can't seem to relate it to $a+b$ and $ab$.

Michael Hardy
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user175533
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5 Answers5

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If $(a,b)=1$, Bezout's Identity says there are $x,y$ so that $ax+by=1$. Then $$ (a^2x+b^2y)(\color{#C00000}{x+y})-(a-b)^2\color{#C00000}{xy}=(ax+by)^2=1 $$ Therefore, $(x+y,xy)=1$.

robjohn
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  • I just found that I answered this question. At least my answer there was different. Looking at the other answers there, I see that mine is very close to Henning Makholm's there. – robjohn Mar 12 '15 at 18:28
  • I was writing a Bezout proof for you as you were writing this - see my Bezout answer here. – Bill Dubuque Mar 12 '15 at 18:37
  • It is pretty close to the answer I mentioned in the comment above. This question should definitely be closed as a duplicate. – robjohn Mar 12 '15 at 18:40
  • Yeah, there's only so many ways to Bezout. I only saw your answer (and link) after I posted mine. There are far too many of these questions to remember them all, and they are not easy to locate by search. – Bill Dubuque Mar 12 '15 at 18:42
  • There are four duplicates mentioned in comments to the question. As long as this question is not deleted, these five questions will remain linked, not just related. – robjohn Mar 12 '15 at 18:44
  • I didn't reload the page since I first opened it, so I didn't see those either. Perhaps there should be dupe notifications too. And oh for a way to search on expressions... – Bill Dubuque Mar 12 '15 at 18:53
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Hint: $a(a+b) - ab = a^2$ and $b(a+b) - ab = b^2$ so that $\mathrm{gcd}(a+b, ab)$ divides $a^2$ and $b^2$.

cactus314
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Since $ab-a(a+b) =-a^2 $ and $ab-b(a+b) =-b^2 $, if $d$ divides both $ab$ and $a+b$, then $d$ divides both $a^2$ and $b^2$.

Since $(a, b) = 1$, $(a^2, b^2) = 1$, so $d = 1$.

marty cohen
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  • @OP Note that it requires proof that $\ (a,b) = 1,\Rightarrow, (a^2,b^2)=1\ \ $ – Bill Dubuque Mar 12 '15 at 18:03
  • The primes dividing a are the ones that divide $a^2$ and similarly for b. – marty cohen Mar 12 '15 at 22:10
  • Which, again, requires proof (e.g. it may fail in some quadratic number rings where irreducibles need not be prime). Probably, for proof, you have in mind the Fundamental Theorem of Arithmetic (existence and uniqunesss of prime factorizations). – Bill Dubuque Mar 12 '15 at 22:15
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Here is a direct proof for those who enjoy Bezout fiddling.

$\quad \begin{align} \gcd(a,b)=1\ \Rightarrow\ 1 &=\, j a + k b\quad\text{for some }\ j,k\in\Bbb Z\\ &=\, (j\!-\!k)a + k(a+b)\\ &=\, (k\!-\!j)b + j(a+b),\ \ \ \text{so multiplying this with above}\\ \Rightarrow\ 1 &=\quad\, m\,ab + n (a+b)\ \ \ \text{for some }\ m,n\in\Bbb Z \\ \Rightarrow\ 1 &=\ \gcd (ab,\ a+b) \end{align}$

Remark $\ $ Using gcd laws eliminates the obfuscatory Bezout coefficients

$$ (a,c)(b,c) = (ab,c(a,b,c)) = (ab,c)\ \ {\rm if}\ \ (a,b,c) = 1\qquad$$

Yours is the special case $\ c = a+b,\ $ and $\ (a,b)=1\ \,(\Rightarrow\, (a,b,c)=1)$

Bill Dubuque
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Hint $\ $ By here: $\ (c,a)=1=(c,b)\,\Rightarrow\, (c,ab)=1.\ $ Let $\ c = a+b.$

Remark $\ $ More generally $\ (a\!+\!b,\, {\rm lcm}(a,b))\, =\, (a,b).\,$ See here for a few proofs.

Bill Dubuque
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