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How would we prove by contradiction that there do not exist integers $x$ and $y$ such that $15x + 1 = 21y$?

I have an idea on how to do this (using a factor of $3$ with $21y = 3(7y)$. I am confused with what I am supposed to do to the other side.

MJD
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ghost
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    Hint: $1=21y-15x=3\times (7y-5x)$ – lulu Oct 18 '22 at 21:33
  • @lulu I appreciate you! Easy question – ghost Oct 18 '22 at 21:49
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    Please, choose the tags carefully, this is not a question about probability nor statistics. – jjagmath Oct 18 '22 at 22:36
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    "Not even a week on this platform and I already have Math Karen's in here " No. Not even a week on this platform and you have people informing you have the expectations, rules, and culture. It seems to me anyone who came into a new place and assumes their behavior is the norm is more deserving of the term "Karen" (although "karen" is a revolting and offensive term and no-one deserves it). – fleablood Oct 18 '22 at 23:13
  • What behaviour? Oh no, I used one too many tags... huge crime I've committed @fleablood But I was just joking, I'll keep this in mind so you can sleep peacefully tonight – ghost Oct 18 '22 at 23:26
  • If so then by first dupe: $,21y-15x = 1\Rightarrow 21,15$ are coprime, contra $,\gcd(21,15)=3\mid 21,15\ $ (this is the easy converse of the Bezout GCD identity). – Bill Dubuque Oct 19 '22 at 11:45

1 Answers1

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By doing this:

$15x-21y=1$. Notice that the left side and the right side share no common divisors (except for 1), which implies that no integers $x$ and $y$ can satisfy the given equation. Alternatively, you can divide both sides by $3$ which gives you $5x-7y$ on the left side and one-third on the right side. But that is a contradiction since, no two integers can add up to a fraction.

冥王 Hades
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