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This question gives one info which is $a\cong b \pmod m$ then we have to conclude that $(a,m)=(b,m)$. Here is what I have so far

$a \cong b \pmod m$ so $a-b=mj$ so $a=mj+b$ , $b=a-mj$

also let $(a,m)=d_1$ and $(b,m)=d_2$ so $ax+my=d_1c$ and $au+mv=d_2k$

I did some algebra and got $$m(jx+y-uj)+bx+au=d_2k+d_1c$$ I am stuck here.... Any hints will be greatly useful

MJD
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Jamie John
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2 Answers2

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$a = b+mj $

Let $d1=\gcd (a,m) $ and $d2=\gcd (b,m)$.

$d1$ divides $m$. And $d1$ divides $a$. So $d1$ divides $a - mj =b$. So $d1$ divides $b$. And as $d1$ divides $m$ we know $d1$ is a common divisor of $b$ and $m$. Now $d2$ is the GREATEST common divisor of $b$ and $m$ so by definition $d2 \ge d1$.

Likewise $d2 $ divides $b+mj=a $ and so $d2$ is a common divisor of $a $ and $m $. So $d1 \ge d2$.

So $d1=d2$.

MJD
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fleablood
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  • Thank you so much!, I followed that d1|a-mj which is d|b but how d1 is a common divisor of b and m ? and how come d2>or equal to d1? I just need little clarification – Jamie John Feb 18 '17 at 03:21
  • @Hanan Ibraheim You know already that $d1$ divides $b$ and by definition of $d1$, $d1$ divides $m$. So, $d1$ is a common divisor of $b$ and $m$. But $d2$ is the gcd of $b$ and $m$ and hence, $d1$ divides $d2$. Because $d1$ and $d2$ are $\geq 1$, we get $d1\leq d2$. – Juniven Acapulco Feb 18 '17 at 03:33
  • so we can use the same linear combination that represents b to say that d1|a-mj , i mean letter wise we used j twice is that fine? I am sorry I am not able to explain what I mean – Jamie John Feb 18 '17 at 03:34
  • By the way, since $d1$ divides $d2$, $$\frac{d2}{d1}=k\in\Bbb N$$ which shows that $$\frac{d2}{d1}\geq 1.$$ – Juniven Acapulco Feb 18 '17 at 03:42
  • @ΘΣΦ GenSan Thank you so much that makes so much sense! – Jamie John Feb 18 '17 at 04:18
  • You are the one who came up with $a = mj +b$ so $b = a- mj$. But the $j$ isnt important. I'd have simply said $a = b + mj$ and $b = a + km$. We can solve for $k$ and discover that $k = -j$ but that's completely irrelevant. – fleablood Feb 18 '17 at 06:21
  • @ΘΣΦGenSan Actually, I was avoiding using the fact that if $k|m,n$ then $k|\gcd(n,m)$ in case that hadn't been a proven result yet. – fleablood Feb 18 '17 at 06:30
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It's easy if you use the following characterization of the greatest common divisor.

$d\ge 0$ is the greatest common divisor of the integers $x$ and $y$ if and only if

  1. $d\mid x$ and $d\mid y$;
  2. for every integer $e\ge0$, if $e\mid x$ and $e\mid y$, then $e\mid d$.

This has the advantage that no hypothesis whether $x$ and $y$ are zero is necessary.

Let $d=\gcd(a,m)$ and write $b=a+mk$. Then $d\mid a$ and $d\mid m$, so clearly $d\mid(a+mk)$, hence $d\mid b$.

Suppose $e\mid b$ and $e\mid m$. Then $e\mid a$, because $a=b-mk$. Therefore $e\mid d$, by the fact $d=\gcd(a,m)$.

Hence $d=\gcd(b,m)$.

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