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Show that if $\gcd(b,c) = 1$ then $\gcd(a,bc) = \gcd(a,b)\gcd(a,c).$

Everyone, I'm stuck on this proof. I know that: $$\gcd(a,bc)=\gcd(a,b)\cdot\gcd(a,c)\gcd(a,bc)\\=\gcd(a,b)\cdot\gcd(a,c),$$ but I don't know how I use $\gcd(b,c)=1$ to get that $\gcd(a,bc)=\gcd(a,b)\cdot\gcd(a,c)$

Michael Rozenberg
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Poonyaphat Mahakaew
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  • If $\gcd(a,b)=1$, we have $\gcd(a,bc)=\gcd(a,c)$ which is true since $a,b$ are coprime, thus they share no common prime factors. In the case $\gcd(a,b)=d\neq 1$, we have $\gcd(a,bc)=\gcd(da_1,db_1c)=d\gcd(a_1,b_1c)$ where $\gcd(a_1,b_1)=1$ and hence by the previous argument, we have $\gcd(a,bc)=d\gcd(a_1,c)=\gcd(a,b)\gcd(a_1,c)$ and since $\gcd(b,c)=1$, they share no common prime factors and thus $\gcd(a_1,c)=\gcd(a,c)$ from which we conclude our result. – Prasun Biswas Aug 27 '17 at 08:15
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    @PrasunBiswas why don't you make that an answer? – Kenny Lau Aug 27 '17 at 08:21

2 Answers2

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Let $a=p_1^{\alpha_1}p_2^{\alpha_2}...p_n^{\alpha_n}$, $b=p_1^{\beta_1}p_2^{\beta_2}...p_n^{\beta_n}$ and $c=p_1^{\gamma_1}p_2^{\gamma_2}...p_n^{\gamma_n}$,

where $p_i$ are different primes and $\alpha_1\geq0$, $\beta_i\geq0$ and $\gamma_i\geq0$ are integer numbers.

Thus, we need to prove that $$\min\{\alpha_i,\beta_i+\gamma_i\}=\min\{\alpha_i,\beta_i\}+\min\{\alpha_i,\gamma_i\},$$ which is obvious because $\beta_i\gamma_i=0$.

Michael Rozenberg
  • 194,933
1

There is another approach just using some basic properties of $\gcd$: \begin{align} \gcd(a,b)\gcd(a,c)&=\gcd(a\gcd(a,c),b\gcd(a,c))\\ &=\gcd(\gcd(a^2,ac),\gcd(ab,bc))\\ &=\gcd(a^2,ac,ab,bc)\\ &=\gcd(a^2,\gcd(ac,ab),bc)\\ &=\gcd(a^2,a\gcd(c,b),bc)\\ &=\gcd(a^2,a,bc)\\ &=\gcd(\gcd(a^2,a),bc)\\ &=\gcd(a,bc)\\ \end{align}

Mercury
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