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Let $N=pq$ for $p,q$ prime.

If $\gcd(a,N)=1$ and $\gcd(b,N)=1$ is it true that $\gcd(ab,N)=1$

That is to say that $a$ and $N$ being coprime, and $b$ and $N$ being coprime, implies that $ab$ and $N$ are coprime.

Could someone give me a simple proof either way please?

lioness99a
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  • That is true for any $N$. What do you already know, Euler's Lemma, Bezout gcd identity, existence and uniqueness of prime factorization (= fundamental theorem of arithmetic)? All can be used to prove it (as well as many other closely related properties). Any way it is a duplicate of many earlier questions.Please search before posting. – Bill Dubuque Dec 22 '16 at 14:52
  • I haven't heard of Euler's Lemma or any of Bezout's Theorems. I know prime factorisation is unique but no more than that. And I did search and didn't find anything useful, hence why I posted... – lioness99a Dec 22 '16 at 14:57
  • It's quite easy using prime factorizations. Where are you stuck?. Btw, I meant Euclid's Lemma above, not Euler's, though one could also use the lesser-know but more general Euler Four Number Lemma. – Bill Dubuque Dec 22 '16 at 16:01

4 Answers4

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Assume by contradiction that $\gcd(ab, N)\neq 1$. Then there exists some prime $p_0$ such that

$$p_0|ab,\ p_0|N$$

However, by Euclid's Lemma, we then have $p_0|a$ or $p_0|b$. That means either

$$p_0|\gcd(a,N)$$

or $$p_0|\gcd(b,N)$$

so one of $\gcd(a,N),\gcd(b,N)$ is not $1$. Thus, we have a contradiction, so our initial assumption is not valid, finishing the proof.

Carl Schildkraut
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    Nice proof. Btw, to the OP, it's not necessary to assume that $N=pq$ because the result is true for every $N$. – Xam Dec 22 '16 at 16:02
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By Bezout's Lemma (see this one), $$ax+Ny=1$$ $$bw+Nz=1$$ for some integers $x,y,z,w$. Simplifying we get $$ab(xw)+N(ybw+axz+Nyz)=1$$ where $xw$ and $ybw+axz+Nyz$ are also integers. This implies that gcd$(ab,N)=1$.

Juniven Acapulco
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$gcd(a,N)=1$ means that $a$ and $N$ have no factor in common (apart from 1). Ditto for $b$. In particular they have no prime factors in common.

Any prime factor of $ab$ is either a factor of $a$ or a factor of $b$ (or both). We know that $N$ has none of these, hence $gcd(ab,N)=1$

user383778
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    You need to justify the claim in your second paragraph. – Bill Dubuque Dec 22 '16 at 15:18
  • That's the definition of primality. It doesn't need justification. – user383778 Dec 22 '16 at 15:51
  • No, that's not the definition of primality in most elementary number theory courses (but it is in more advanced contexts). Rather, it is a consequence of other basic named results, and one should explicitly state which of those results is being used. – Bill Dubuque Dec 22 '16 at 15:55
  • Sorry - it is the definition of primality, and it is not a very advanced concept. It's a shame that elementary number theory fixes the mind on the notion of irreducibility as a placeholder for primality, and it causes no end of bother when students advance to studying ring theory. – user383778 Dec 22 '16 at 15:59
  • Apparently you did not even bother to read what I wrote. – Bill Dubuque Dec 22 '16 at 16:03
  • Yes I did - and while I'm here in your answer you directly invoked the fundamental theorem of arithmetic, which relies on the definition of primality for its proof. That's a seriously powerful tool you've used, so why get high and mighty with me for using a more basic construct? – user383778 Dec 22 '16 at 16:08
  • The issue above is independent of other answers (and if you read the comments on the question you will learn that I used that approach because it was the only one the OP mentioned she was familiar with). In any case, if you are going to use more advanced concepts / terminology then it is essential to explain that, else it could prove very confusing or misleading. And if that was you who downvoted my answer, then it would be very interesting to learn why. – Bill Dubuque Dec 22 '16 at 16:24
  • Also, minor cosmetic detail: \gcd(a, N) gets you $\gcd(a, N)$. – Mr. Brooks Dec 22 '16 at 21:56
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Hint $\ $ By the uniquness of prime factorizations, the prime factorization of $ab$ is that obtained by concatenating the unique prime factorizations of $\,a\,$ and $\,b.\,$ The hypotheses imply that the primes $\,p,q\,$ do not occur in either so they do not occur in their concatenation.

Bill Dubuque
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