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If $a\mid bc$ and $\gcd(a,b)=1$, prove that $a\mid c$ without using Bezout's Identity.

Maht Tomato
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3 Answers3

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Denote $\mathrm{gcd}(a,b) =: (a,b)$. Then $a\mid bc$ and $a\mid ac$ implies $a\mid (ac,bc)$ i.e $a\mid c(a,b)$.

AlvinL
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  • That seems to be the beg the question. Not sure if we know that "all divisors divide $\gcd$" or $\gcd(ma,mb) = m\gcd(a,b)$ without also knowing what we are trying to prove. Even if it's not completely circular it does seem in essence to require proving the idea. – fleablood Oct 25 '19 at 16:23
  • @fleablood It need not be circular since the gcd distributive law can be proved without using Bezout. Indeed, it holds in any gcd domain, which needn't be Bezout, e.g. $,\Bbb Z[x], $ and $,\Bbb Q[x,y].\ \ \ $ – Bill Dubuque Oct 25 '19 at 18:42
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Do strong induction on $b$. The case $b=1$ is obvious.

Suppose the statement holds for all numbers less than $b>1$. Take $p$ a prime divisor of $b$ and write $b=px$. Then $a\mid x(pc)$, so $a\mid pc$ by the induction hypothesis. In particular $pc=ay$ and, by Euclid’s theorem, either $p\mid a$ or $p\mid y$. The former case would imply $p\mid\gcd(a,b)$: contradiction. Hence $p\mid y$, $y=pz$ and $pc=apz$, so $a\mid c$.

The proof above uses (1) every integer $>1$ is divisible by at least a prime; (2) Euclids’s theorem if $p\mid xy$, for prime $p$, then either $p\mid x$ or $p\mid y$.

From these two facts one can deduce unique factorization into primes. Conversely, uniqueness of factorization into primes implies both (1) and (2) above.

On the other hand, the statement under scrutiny implies Euclid’s theorem. Assume it and suppose $p$ is a prime and $p\mid xy$. If $p\nmid x$, then $\gcd(p,x)=1$, so $p\mid y$ by assumption.

Here $\gcd(A,B)=1$ can be considered as a shorthand for “no prime divides both $A$ and $B$”, we don't actually need existence of the gcd in all the arguments above.

egreg
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First, show by strong induction that every integer greater than $1$ has a prime factorization. If $a | bc$ and $\gcd(a,b)=1,$ then $bc = ka,k\in\mathbb{Z}$ and $a$ and $b$ do not have any prime factors in common. Since $bc=ka,$ $k$ must contain all the prime factors to powers equal to those of $b$ (this applies even if $b$ has no prime factors), so $\frac{ka}{b} = c$ must be an integer. Since this is an integer multiple of $a,$ by definition $c | a.$

Simon Fraser
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    "k must contain all the prime factors of b " You should put is something about $k$ contains all the prime factors to powers equal to those of $b$. It's not enough to so $30$ contains all the prime factors of $12$ (namely $2$ and $3$) so $12|30$. You must say if $p^m|b$ and so $p^m|ka$ but $p\not \mid a$ so $p^m|k$. And thus $b|k$. – fleablood Oct 25 '19 at 08:14
  • Okay @fleablood that was clearly a major mistake. I guess just looking for counterexamples is a really nice way of checking if I'm right. Thanks for the insight! – Simon Fraser Oct 25 '19 at 14:08
  • I wouldn't call it a major mistake. I'm not even sure I'd call it a mistake at all. The point is as $a$ and $b$ have no factors in common the the "components" of $a$ must be "contained in" $c$. That is, IMO, best way (most direct) to do it and how I'd do it. But the exact wording and justification needs a teensy tad of care. If we assume unique factorization has been proven you can do a prime by prime deconstruction. IMO it feels heavy-handed but it is airtight and complete. – fleablood Oct 25 '19 at 16:19