0

Let $(ca)x + (cb)y = cd$ where $d = (a, b).$

Then since $\gcd(ca,cb)\mid ca$ and $\gcd(ca,cb)\mid cb \to \gcd(ca,cb)\mid cd$.

I don't get how they deduced the conclusion. For one thing, $\gcd(ca,cb)\mid ca$ and $ \gcd(ca,cb)\mid cb \to \gcd(ca,cb)\mid (c(a + b)) $ where $a + b = d$ only if $a + b = ax + by$ meaning $x = y = 1$. Please, elaborate on this.

Community
  • 1
user262122
  • 85

1 Answers1

0

I can't make head or tails of your objection starting with "For one thing..." It is true that $\gcd(ac,bc)\mid c(a+b)$. Why is that a problem?

You are making this much harder than it is. If $D\mid A$ and $D\mid B$ then $D\mid Ax+By$ for any integers $x,y$.

Now let $A=ac, B=bc,$ and $D=\gcd(A,B)=\gcd(ac,bc)$.

Thomas Andrews
  • 177,126
  • So, there are more than one pair of solutions for $(x, y)$ here? Possibly infinite number of solution pairs? – user262122 Aug 18 '15 at 01:21
  • Please, be very specific. "here?" $x,y$ are solutions to $(ac)x+(bc)y=\gcd(ac,bc)$. Presumably, you've shown that such $x,y$ exist. And yes, if there is a solution to $Ax+By=D$, there are infinitely many, but I'm pretty sure that's not what you mean. – Thomas Andrews Aug 18 '15 at 01:25
  • How would $D|A, D|B \to Ax + By$ for some $x, y$ be different from $D|A, D|B \to Ax + By$ for any $x, y$ ? Also, you guessed right, I asked exactly the question about the number of solution pairs to $Ax + By = D$. – user262122 Aug 18 '15 at 01:35
  • Do you mean $D\mid A,D\mid B\rightarrow D\mid Ax+By$? Because $\rightarrow Ax+By$ doesn't mean anything. $Ax+By=\gcd(A,B)$ for some integers $x,y$. "For any" or "for all" means that you can choose any $x,y$ and it is still true. So $D\mid A,D\mid B\rightarrow D\mid Ax+By$ for any integers $x,y$. – Thomas Andrews Aug 18 '15 at 01:44
  • Oh, ok. I see. Thanks. – user262122 Aug 18 '15 at 01:46