0
$\begingroup$

Let $m$ be a natural number larger than 1, and suppose that $m$ satisfies the following property:

For any integers $a$ and $b$, if $m$ divides $ab$, then $m$ divides either $a$ or $b$ (or both).

Show that $m$ must be prime.

I think we should solve this question with a proof by contradiction. So, suppose $m$ is not prime, therefore $\exists$$k$ $\in$ $\mathbb{Z}$ such that $k|m$. We also notice that the property $m$ satisfies is Euclid's Lemma, and from it we can conclude $gcd(a,m) = m$ and $gcd(b,m) = m$ depending on if $m|a$ or $m|b$ or both.

Now from here I don't know how I should proceed further. In fact, I'm not sure if I'm looking in the right direction, so your help would be appreciated.

$\endgroup$
  • $\begingroup$ Hint: If $m$ is not prime then $m=ab$ with $m>a,b>1$. $\endgroup$ – lulu Jul 20 '19 at 19:49
0
$\begingroup$

You are right: suppose $m\in \mathbb{N}$ is not prime and let's fix $(k,l)\in \mathbb{N}^2$ such that $m=kl$ and $ k,l \neq m$ then either $m\mid k$ or $ m\mid l $ which is absurd as both k and l are strictly less than m (and positive)

$\endgroup$
0
$\begingroup$

Hint: $ $ first show: $\,p\,$ is prime $\!\iff\! [\,p \color{#c00}{\bf =} ab\,\Rightarrow\, p\mid a\,$ or $\,p\mid b\,],\,$ then use $\,p\color{#c00}{\bf =}ab\,\Rightarrow\, p\mid ab$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.