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I have below two statemetns :

$$\gcd(a,b) = \gcd(r, b)$$ $$\gcd(a,b) = \gcd(a\bmod b, b)$$

Is there any difference between them ?

Michael Hardy
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rrr
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  • I depends on what $r$ is. Is there any more context? – Bart Michels Sep 02 '14 at 21:40
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    I'm having trouble understanding the question. Sometimes (perhaps especially in recent decades, since 1945 or so?) $a\bmod b$ is taken to mean the remainder on division of $a$ by $b$. If that's what it means, then the two statements say the same thing only if $r=(a\bmod b)$. But even if $r\ne(a\bmod b)$, they would be equivalent in another sense: deducible from each other. However, in your subject line you've written "$\equiv$". That is used when the older sense of "mod" is intended where $a\equiv r\bmod b$ is parsed as $(a\equiv r)\bmod b$, and means that $a$ and $r$ both leave the.... – Michael Hardy Sep 02 '14 at 21:42
  • ....same remainder on division by $b$. ${}\qquad{}$ – Michael Hardy Sep 02 '14 at 21:42
  • Sorry I am new to number theory and I really don't have any particular preference on notation yet. I am just going through this post and trying to understand the replies http://math.stackexchange.com/questions/895059/is-gcda-b-gcda-mod-b-b/895088#895088 – rrr Sep 02 '14 at 21:57
  • In that link, they are using $r$ and $a\mod b$ as same thing. From your reply it seems they both are different and not always equal as the actual congruence definition is : $$a\equiv b \mod n \iff n|(a-b)$$ Have I understood your reply correctly ? – rrr Sep 02 '14 at 21:58
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    depand on your question ,both statments are equivalent , i mean if you have $a\equiv r \bmod b$ then $gcd(r,b)=gcd(a \bmod b ,b ) \iff gcd(a,b)=gcd(r,b)$ – Bswan Sep 02 '14 at 21:59
  • @Bswan So is it acceptable to replace $r$ by $a \mod b $ if the context in which it is used is clear enough ? (as in the link provided above) – rrr Sep 02 '14 at 22:03
  • gcd(a,b)=gcd(r,b)=d , then a=dn ,b=dm s.t gcd(n,m)=1 however r=dl and gcd(l,m)=1 , so in this case its ok you can replace a with r in mod condition. – Bswan Sep 02 '14 at 22:03
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    IMHO its enough to show for r=dl and b= dm , gcd(m,l)=1 , that means what ever a,r are , they would only share d with b.gcd(a,r,b)=d ( so nothing wrong with replacing ) – Bswan Sep 02 '14 at 22:12

1 Answers1

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The proof below shows: $\ \color{#c00}{a\equiv r\pmod b}\ \Rightarrow\ \gcd(a,b) = \gcd(r,b).\,$

In particular it's true for $\ r = (a\ {\rm mod}\ b),\ $ i.e. the least $\,r\ge 0\,$ such that $\ a\equiv r\pmod b$.

Therefore your second statement is a special case of your first statement.

More generally $\,a\equiv r\pmod b\,\Rightarrow\, \gcd(f(a),b) = \gcd(f(r),b)\,$ for any polynomial $\,f(x)\,$ with integer coefficients, because $\,f(a)\equiv f(r)\pmod b\,$ by the Polynomial Congruence Rule.

Proof $\ $ If $\ d\mid b\ $ then $\ d\mid a\!\iff\! d\mid r,\ $ since $\,d\mid \color{#c00}{b\mid a\!-\!r}.\ $ Therefore $\,a,b\,$ and $\,r,b\,$ have the same set $\,S\,$ of common divisors $\,d,\,$ so they have the same greatest common divisor $(= {\rm max}\,S)$.

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Bill Dubuque
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