5

Let $\gcd (a, b) = d$. So, $ax + by = d$ for some $x, y$. Then $(ca)x + (cb)y = cd$. Thus, $\gcd (ca, cb) = cd = \gcd(a, b)c$.

Does it work?

Martin Sleziak
  • 53,687
user262122
  • 85

2 Answers2

0

No. The problem is that $ax+by = d$ does not imply $\gcd(a,b) = d$.

The statement $\gcd(ca,cb) = c\gcd(a,b)$ is true, however. To see that note that

$\gcd(a,b) \mid a \wedge \gcd(a,b) \mid b \implies c\gcd(a,b) \mid ac \wedge c\gcd(a,b) \mid bc$ so $$\gcd(ac,bc) \mid c\gcd(a,b)$$

The reverse is a little bit more work (you need $\gcd(ac,bc) \mid c\gcd(a,b)$ to conclude).

AlexR
  • 24,905
  • gcd(a,b) can also be defined as min ( { ax+by | for all x,y } ) – RowanS Aug 17 '15 at 20:35
  • @RowanS For all $x,y$ in what set? I can make your quantity arbitrarily small for $x,y\in\mathbb Z$... – AlexR Aug 17 '15 at 20:38
  • @RowanS No, that doesn't work, because that set doesn't have a minimum. We can choose $x$ and $y$ to make that set get closer and closer to $-\infty$. – Ben Sheller Aug 17 '15 at 20:38
  • 1
    @AlexR gcd(a,b) can be defined as min({ax+by| for all x,y in z and ax+by>0)} so therefore cgcd(a,b) = cmin({ax+by| for all x,y in z and ax+by>0)} )=min({cax+cby| for all x,y in z and ax+by>0)} ) – RowanS Aug 17 '15 at 20:41
  • @ AlexR, does the following chain of implications hold? Let $c\gcd(a, b) | bc$. Since $\gcd(ac, bc) | (bc)$, $\gcd(ac, bc) \to c\gcd(a,b)$? – user262122 Aug 17 '15 at 21:07
  • @user262122 No, $x\mid z$and $y\mid z$ says nothing about the relation between $x$ and $y$. – AlexR Aug 17 '15 at 21:09
  • @ AlexR, $(ac, bc) | ac, (ac, bc) | bc \to (ac, bc) | c(a, b)$ because $c(a, b)| ac, c(a, b) | bc?$ – user262122 Aug 17 '15 at 21:34
  • @user262122 Nope. That's the same problem as before. – AlexR Aug 17 '15 at 21:35
  • @ AlexR can you elaborate on the conclusion in your post. I am having hard time seeing it. – user262122 Aug 17 '15 at 22:23
  • @AlexR your first line is wrong. user2621222 first states let gcd(a,b)=d then considers ax+by = d and nowhere implies that if ax+by=d then gcd(a,b)=d. – RowanS Aug 18 '15 at 00:15
0

You can use prime factorization if you don't want to work through cumbersome axioms about what properties GCDs satisfy, or work with formulas expressing GCDs of different integers using linear combinations. If you accept prime factorization, your statement is basically trivial. Maybe that means you're not supposed to solve it that way though ;)

user2566092
  • 26,142