Elementary number theory, concerning GCD

Question

If I have two co-prime integers. $a, b$. Suppose that the product of these two integers is $c$.

Further suppose that I have a further product of two co-primes, so $d = af$. Now if I multiply these together to obtain:

$$cd = a^2bf.$$

From this i then say that $\gcd(cd, a^4) = a^2$.

Its clear that $a^2|a^4$ and $a^2|cd$.

Why can't the $\gcd = a^3$?

What I can't get my head around is that I know $a, b$ and $a, f$ are coprime (from this I can't deduce that the three integers are mutually coprime, can I? ), but surely there must exist a product $bf$ which has factor $a$?

My idea is clearly flawed otherwise $a^3$ would be the answer. I really would like to put this idea to rest.

$\gcd(cd, a^4 ) = \gcd (a^2bf, a^4) = a^2 \gcd (bf, a^2) = a^2 .1$ — anumosh, Mar 06 '15 at 21:27
$a,b,f$ need not be mutually coprime. $a$ and $bf$ are coprime. — anumosh, Mar 06 '15 at 21:29

score 1 · Accepted Answer · answered Mar 06 '15 at 21:29

1

Hint: By the Bezout theorem show that if $gcd(a,b)=1$ and $gcd(a,f)=1$ then $gcd(a,bf)=1$.

answered Mar 06 '15 at 21:29

Fermat

5,230

score 0 · Answer 2 · edited Apr 13 '17 at 12:21

0

Notice $\ (a^4,a^2bf) = a^2\color{#c00}{(a^2,bf)}\,$ by the gcd Distributive Law.

But $\,(a,b)=1=(a,f)\,\Rightarrow\,(a,bf) = 1\,$ by Euclid's Lemma.

Again $\,(a,bf)=1=(a,bf)\,\Rightarrow\,\color{#c00}{(a^2,bf)=1}\,$ by Euclid's Lemma.

Hence $\ (a^4,a^2bf) = a^2.$

edited Apr 13 '17 at 12:21

Community

1

answered Mar 06 '15 at 21:40

Bill Dubuque

272,048

Elementary number theory, concerning GCD

2 Answers2