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Let $m,a_1,\dots,a_k\in\mathbb{N},$ show that: $$m\mid \gcd(a_1,\dots,a_k)\text{ implies } \gcd(a_1,\dots,a_k)= m\gcd(\frac{a_1}{m}, \dots,\frac{a_k}{m}), \gcd\left(\frac{a_1}{\gcd(a_1,\dots,a_k)},\dots,\frac{a_k}{\gcd(a_1,\dots,a_k)}\right)=1$$

My Attempts:

Since $m\mid \gcd(a_1,\dots,a_k)$ and $\gcd(a_1,\dots,a_k)\mid a_j$ where $j\in[1,k]\cap\mathbb{N}$ that $m\mid a_j$ we have: \begin{align} \frac{\gcd(a_1,\dots,a_k)}{m}\mid \frac{a_j}{m} &\equiv \frac{\gcd(a_1,\dots,a_k)}{m}\mid \frac{a_1}{m},\cdots,\frac{\gcd(a_1,\dots,a_k)}{m}\mid\frac{a_k}{m}\\ &\Rightarrow \frac{\gcd(a_1,\dots,a_k)}{m}\le \gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\\ &\Rightarrow \gcd(a_1,\dots,a_k)\le m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\tag{1} \end{align} Similarly we have: \begin{align} \gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\mid \frac{a_j}{m} &\equiv \gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\mid \frac{a_1}{m},\cdots,\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\mid\frac{a_k}{m}\\ &\equiv m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\mid a_1,\cdots,m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\mid a_k\\ &\Rightarrow m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\le\gcd(a_1,\cdots,a_k)\\ &\Rightarrow m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\le\gcd(a_1,\dots,a_k)\tag{2} \end{align}

Put $(1)$ and $(2)$ together that:

$$m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})\le\gcd(a_1,\dots,a_k)\le m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})$$

That $\gcd(a_1,\dots,a_k)= m\gcd(\frac{a_1}{m},\dots,\frac{a_k}{m})$
Let $m=\gcd(a_1,\dots,a_k)$ we have: $$\gcd(a_1,\dots,a_k)\gcd(\frac{a_1}{\gcd(a_1,\dots,a_k)},\dots,\frac{a_k}{\gcd(a_1,\dots,a_k)})=\gcd(a_1,\dots,a_k)$$ Divide both side by $\gcd(a_1,\dots,a_k)$, proved $\gcd\left(\frac{a_1}{\gcd(a_1,\dots,a_k)},\dots,\frac{a_k}{\gcd(a_1,\dots,a_k)}\right)=1\tag*{$\square$}$

Is my proof correct, and are there better approaches $?$

Thanks for your help.

Andronicus
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Ethan
  • 5,291

2 Answers2

3

Using inequalities seems like overkill.

As a general property, we have and will use $$m\mid \gcd(a_1,\dots,a_k) \hspace{0.2cm}\Leftrightarrow \hspace{0.2cm} m \mid a_{p}, \forall p\in\{1..k\} \hspace{2cm} (1)$$

Assume $\gcd\left(\frac{a_1}{\gcd(a_1,\dots,a_k)},\dots,\frac{a_k}{\gcd(a_1,\dots,a_k)}\right)=n$ .

Then by $(1)$ we get $n \mid \frac{a_p}{\gcd(a_1,\dots,a_k)}, \forall p\in\{1..k\}$.

This can be written as $(n \gcd(a_1,\dots,a_k))\mid a_{p},\forall p\in\{1..k\}$

By $(1)$ we get $(n \gcd(a_1,\dots,a_k))| \gcd(a_1,\dots,a_k)$ .

From this we get $n\in \{-1,+1\}$, but $n$ is pozitive so $n=1$.

user3257842
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2

This implication does not make sense, because $m$ is never used. But the implication is always true. Suppose:

$$\gcd\left(\frac{a_1}{\gcd(a_1,\dots,a_k)},\dots,\frac{a_k}{\gcd(a_1,\dots,a_k)}\right)=n$$

This would mean, that $\frac{a_i}{\gcd(a_1,\dots,a_k)}$ is always divisible by $n$ - contraddiction, because in denominator is $\gcd(a_1,\dots,a_k)$, so there cannot be any common divisors.

Andronicus
  • 3,436