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How can I prove this by using contradiction?

Given the premises, assume that there does exists an integer $b$ such that $ab \equiv 1 \pmod m.$

Then, let $k = \frac{ab - 1}{m},$ where $k \in \mathbb Z$.

So, $$ab = km + 1, b = \frac{km+1}{a} = \frac{km}a + \frac 1a.$$

Then I try to prove $km/a$ is an integer since I know $1/a$ is not.

amWhy
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DDD
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    Start by assuming the premise and assuming there does exist an integer $b$ such that $ab\equiv 1\pmod m$. Then derive a contradiction. – amWhy Apr 03 '20 at 17:35
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    That's how you prove, after deriving a contradiction, that there doesn't exist an integer $b$ such that .... Glad I could answer your question. – amWhy Apr 03 '20 at 17:41
  • Thank you for your reply! I did what you said and got b= (km+1)/a = km/a + 1/a. But I stuck on that because I don't know how to do with km/a. – DDD Apr 03 '20 at 17:41
  • Please update you post with your proof in progress. – amWhy Apr 03 '20 at 17:43
  • I have uploaded it. – DDD Apr 03 '20 at 17:47
  • Thanks for your cooperation! I formatted the post to make it "prettier". Note that the premise informs us that $\gcd(a, m)>1$. – amWhy Apr 03 '20 at 17:51
  • But I think gcd cannot imply if m is a multiple of a... Sorry I don't get your point. – DDD Apr 03 '20 at 17:57
  • Try Bezout's identity "Let a and b be integers with greatest common divisor d. Then, there exist integers x and y such that ax + by = d." – Somos Apr 03 '20 at 18:02
  • @Alan Your current strategy is too narrow for success, but you’re on the right track. It’s not reasonable to expect $km/a$ to be an integer, but we can show that $km/d$ to be an integer, where $d = \gcd(a,m)$. – Erick Wong Apr 03 '20 at 18:04
  • I tried to use Bezout's identity to represent m and a, but it is still a mess...should I focus on the whole function instead of km/a? – DDD Apr 03 '20 at 18:20
  • @ErickWong Thank you for your advice. But in my understanding, if we prove km/d is an integer, it still leaves another integer serving from Bezout's identity. – DDD Apr 03 '20 at 18:42
  • You can also use the contrapositive. If $ab \equiv 1 \pmod{m}$ for some $b$ then $\gcd(m,a)=1$. – Alan Muniz Apr 03 '20 at 19:51
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    Hint $ $ if $,d\mid a,m,$ then $,d\mid ab!-!km = 1.\ $ $\phantom{}$ See here for the general result. More generally if $,\gcd(a,m)>1,$ then $,a,$ is a zero-divisor so not a unit (invertible) – Bill Dubuque Apr 03 '20 at 19:57
  • @DDD No, I’m saying do exactly as your current proof but instead of looking at $(km+1)/a$, look at $(km+1)/d$, which is a multiple of $b$ and hence also an integer. – Erick Wong Apr 03 '20 at 20:44

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