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I'm getting started with proofs but I have no idea how to do this. I have gotten to $b c = a k$ where $k$ is some integer by divisibility. I'm not sure I can use division yet as they will no longer be integers. I have no idea how to prove it and am seeking helpful ideas.

Bill Dubuque
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Alexander
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  • Hint $\bmod a!:\ bc,$ differs from $,b,$ by a $\color{#0a0}{{\rm unit\ multiple} \ c,},$ so if one is $\color{#c00}{\equiv 0},$ then so too is the other, explicitly: $$\large \bmod a!:\ \color{#0a0}{cs\equiv 1},\ {\rm so},\ b\equiv b(\color{#0a0}{cs})\equiv (\color{#C00}{bc})s\equiv 0,,\ {\rm by}\ ,\color{#C00}{bc\equiv 0},\ {\rm by},\ a\mid bc $$ – Bill Dubuque Feb 10 '21 at 02:24
  • $\large {\rm i.e.\ scale},\ bc\equiv 0,\ {\rm by},\ \color{#0a0}{c^{-1}\equiv s},\ {\rm to\ get},\ b\equiv 0\ \ $ – Bill Dubuque Feb 10 '21 at 02:38
  • I got the answer and apparently does not make use of congruence. I explained the answer on the comments to the answer below. Unfortunately, this has been associated with another question that is not similar at all to be considered duplicate and cannot answer it. Appreciate the help. – Alexander Feb 11 '21 at 07:48
  • That's the same as the method in the first linked dupe, i.e. scale the Bezout equation by $,b.,$ But the above congruence method is more conceptual arithmetically, showing that it is a special case of the ubiquitous fact scaling an equation (or congruence) by an invertible yields an equivalent congruence. Generally equations (on arithmetical operations) are much simpler to reason with than are divisibility relations. So the sooner you learn the (arithmetical) language of congruences (modular arithmetic) the simpler number theory will be. – Bill Dubuque Feb 11 '21 at 08:05

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Hints

Suppose that $1 = ar + cs$ and suppose that $a,c$ have a common factor $d$. Can you prove that therefore $d$ must divide $1$?

Suppose that $a,c$ are relatively prime and that $a$ divides $(bc)$. Can you prove that therefore $a$ divides $b$?

Have you been taught about the prime factorization theorem?

If $a,c$ are relatively prime, is it possible for any prime number to appear in both the prime factorization of $a$ and the prime factorization of $c$?

If $a$ divides $(bc)$ then how does the prime factorization of $a$ compare with the prime factorization of $(bc)$? What conclusion can you then draw, under the assumption that the prime factorizations of $a$ and $c$ are completely disjoint?

user2661923
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  • We have not been using primes or anything like that yet for proofs but I do know prime factorization – Alexander Feb 10 '21 at 03:20
  • @Alexander You have to find some way of proving that if $a$ and $c$ are relatively prime and $a|(bc)$, then $a|b.$ There may be some clever alternative to proving and using this assertion, but I can't think what it might be. – user2661923 Feb 10 '21 at 03:22
  • I actually did not have to apparently I had to use the 1=ar+cs and multiply both sides by b resulting in 1 = arb + bcs and since bc = ak it could be replaced. After that I could factor out 'a' and by divisibility it a|b. As explained we are not using any fancy math just yet. Appreciate the help. – Alexander Feb 11 '21 at 07:38
  • @Alexander That's exactly the same as the common proof in the first linked dupe. Did you not even follows the dupe links? See my comment on the question for further remarks. – Bill Dubuque Feb 11 '21 at 08:11