$c\mid ab\Rightarrow c\mid \gcd(a,c)\gcd(b,c),\,$ for integers $a, b, c$

Question

Show that if $a, b$ and $c$ are integers with $c|ab$ then $c|(a,c)(b,c)$.

Now $(a, c)$ and $(b, c)$ would both divide $c$ since it's the gcd, but how would I show $c$ divides their product, and $(a,c)(b, c) \geq c$ and not the other way around?

Answer 1

$(a,c) = ax_1 + cx_2$ and $(b,c) = bx_3 + cx_4$ for some $x_1, x_2, x_3, x_4$.

Thus $(a,c)(b,c) = (ax_1+ cx_2)(bx_3 + cx_4) = abx_1x_2 + cx_2bx_3 + cx_4ax_1 + c^2x_2x_4$. Since $c$ divides $ab$, it divides each of the 4 terms in the expansion so it divides $(a,c)(b,c)$.

(Remark: The expansion also shows that $(a,c)(b,c) \geq c$.)

Answer 2

Hint $\rm\ \ c\mid ab\ \Rightarrow\ c\mid (ab,bc,ac,cc)=(a,c)(b,c)\ $ by the distributive law for gcd.

More explicitly $\rm\ (a,c)(b,c) = ((a,c)b,(a,c)c) = (ab,cb,ac,cc)\ $ by distributing a few times (and implicitly using associativity of gcd). See this answer for more on gcd arithmetic

Remark $\ $ If you write the gcd's in Bezout linear form, then the above proof becomes essentially the more specific proof in Kelvin's answer. Then the distributive law for gcds becomes simply the distributive law for integers. But the proof becomes less general, depending on the fact that gcds have linear representation (Bezout's Lemma), which is not always true in more general domains, e.g. the above proof still works in the polynomial rings $\,\Bbb Z[x]\,$ and $\,\Bbb Q[x,y]\,$ where Bezout fails, e.g. $\,x,y\,$ are coprime but $\, ax+by= 1\Rightarrow 0 = 1$ by eval at $\,x=0=y).$

Answer 3

Hint: for a given prime $p$, consider what powers of $p$ divide each of these.