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If $ b\equiv c \pmod m $ then prove that $(b,m)=(c,m)$.

My solution so far:

If $b \equiv c \pmod m$ then $c = b - sm$ for some $s$. Now if $d = (b, m)$ then $ d | (b + s m)$.

Bill Dubuque
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jacksparrow007
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4 Answers4

8

Hint $\ $ If $\rm\,d\:|\:m\:$ then $\rm\:d\:|\:b\!\iff\! d\:|\:c,\:$ by $\rm\:b\equiv c\pmod{\! d},\,$ by $\rm\,d\:|\:m\:|\:b\!-\!c$

Therefore $\rm\,m,b,\,$ and $\rm\,m,c\,$ have the same set $\,\rm S\,$ of common divisors $\rm\,d,\,$ so they necessarily have the same greatest common divisor $\rm (= \max\:\! S)$

Remark $\ $Alternatively, we may use the Bezout characterization of the gcd $\rm\,(m,b)\,$ as the least positive integer in $\rm\: m\,\Bbb Z + b\,\Bbb Z. \,$ Thus $\rm\:(m,b) = (m,c)\:$ follows from $\rm\: m\,\Bbb Z + b\,\Bbb Z\, =\, m\,\Bbb Z + c\,\Bbb Z,\,$
true by $\rm\ m\:j+ \color{#c00}bk = m\:j + (\color{#c00}{b\!-\!c})k + \color{#c00}ck = m\,(j+k(b\!-\!c)/m) + ck = m j' + ck,\:$ so $\rm\:m\,\Bbb Z + b\,\Bbb Z \,\subseteq\, m\,\Bbb Z + c\,\Bbb Z,\:$ and the reverse inclusion follows by $\rm\,b\leftrightarrow c\,$ symmetry.

In terms of ideal theory this may be viewed simply as a standard "change of basis", cf. Remark here.

Bill Dubuque
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  • $mj + (b-c)k + ck = m(j + k(b-c)/m) + ck$. How do you reach the right side of this? – Michael Munta Aug 26 '19 at 13:32
  • Also how to interpret this $\rm:m,\Bbb Z + b,\Bbb Z \subset m,\Bbb Z + c,\Bbb Z.:$? – Michael Munta Aug 26 '19 at 13:35
  • @Michael That inclusion means every integer of the form $\rm,m,j+b,k,\ \ j,k\in\Bbb Z,$ can be written as $\rm,m,j'+c,k',\ \ j',k'\in\Bbb Z.,$ The rhs arises by factoring $\rm m$ out of $\rm,b!-!c,,$ i.e. $\rm, (b!-!c)k = m(k(b!-!c)/m)$ $\ \ \ \ $ $\ \ \ \ $ – Bill Dubuque Aug 26 '19 at 14:53
  • Ok, but specifically how to interpret the proper subset notation? – Michael Munta Aug 26 '19 at 15:02
  • @Michael $\ \rm m,\Bbb Z + b,\Bbb Z := { m,j + b,k\ :\ j,k\in\Bbb Z},$ denotes a set so it is normal set inclusion notation. – Bill Dubuque Aug 26 '19 at 15:31
  • But $\subset$ implies that the left set has less elements than the right one. By which logic? – Michael Munta Aug 26 '19 at 16:15
  • @Michael Not for infinite sets, e.g. $, 2\Bbb Z\subsetneq \Bbb Z\ $ but both have the same size (cardinality). – Bill Dubuque Aug 26 '19 at 16:26
  • I cant seem to wrap my head around this. Can you please show an example of this proper subset with actual numbers for this specific case? – Michael Munta Aug 26 '19 at 19:17
  • @Michael e.g. above the even integers are a proper subset of the integers but they have the same cardinality. If you're not familiar with cardinality of (infinite) sets then there are surely many explanations in answers on MSE. e.g. here,. – Bill Dubuque Aug 26 '19 at 19:28
  • I know about that, but I meant for you to show the one from the answer with linear combination. Surely the left set will have elements that are not part of the right set, or? – Michael Munta Aug 26 '19 at 19:37
  • I mean if any integer that can be written in the first form can also be written in the second then those sets contain exactly the same integers, right? So what makes the first set a proper subset of second, where is the difference? – Michael Munta Aug 26 '19 at 20:04
  • @Michael No it means that all those of first form are a subset of all those of 2nd form. It seems you are interpreting $\ S\subset T,$ in the answer as denoting a proper subset. But there it only means subset (the notation is used both ways). I'll edit it to remove that ambiguity. – Bill Dubuque Aug 26 '19 at 20:23
  • So those 2 sets are in fact the same? – Michael Munta Aug 27 '19 at 16:26
  • @Michael Yes, the answer proves one inclusion and the reverse inclusion follows by symmetry. Thus they have the same least positive element (= gcd). In fact both $\rm = d,\Bbb Z,$ where $\rm,d = (b,m) = (c,m)\ \ $ – Bill Dubuque Aug 27 '19 at 16:33
6

Suppose that $b\equiv c\pmod{m}$.

Suppose that $x$ divides both $b$ and $m$. Since $c=b+km$ for some $k$, it follows that $x$ divides $c$.

Similarly, if $x$ divides both $c$ and $m$, then $x$ divides $b$.

Thus every common divisor of $b$ and $m$ is a common divisor of $c$ and $m$, and vice-versa.

It follows that the greatest common divisor of $b$ and $m$ is the same as the greatest common divisor of $c$ and $m$.

André Nicolas
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3

You pointed out that $b=c+sm$ for some $s$.

Let $d=bx+my$ for some $x,y\in \mathbb{Z},\;\;d\in \mathbb{N}$. Then $d=cx'+my'$ for $x'=x,\;\;y'=y+sx$. Similarly if $d=cx+my$ for some $x,y\in \mathbb{Z},\;\;d\in \mathbb{N}$ then $d=bx'+my'$ for $x'=x,\;\;y'=y-sx$.

Bearing in mind that $(b,m)=\min\{bx+my:x,y\in \mathbb{Z},bx+my>0\}$ the result follows.

P..
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2

Let $(b,m) = d$ and $(c,m) = d'$. If $b \equiv c \pmod m$ then $b = c + \lambda m$ for some $\lambda \in \mathbb{Z}$. We know that $d|b$ and $d|m$, so $d|(b-\lambda m)$ and so $d|c$ and hence $d|d'$.

Can you see where to take it from here?

Clive Newstead
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