Looking at the following one would think it is simple, and mabye it is, BUT to me it is impossible and has been so for two days.

So, show that if x≡y (mod a) then (x,a)=(y,a)

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- Jan 30th 2010, 04:55 AMmatzerathx≡y (mod a) => (x,a)=(y,a)
Looking at the following one would think it is simple, and mabye it is, BUT to me it is impossible and has been so for two days.

So, show that if x≡y (mod a) then (x,a)=(y,a) - Jan 30th 2010, 05:45 AMSudharaka
Dear matzerath,

Suppose, $\displaystyle d_1=(x,a)$ and $\displaystyle d_2=(y,a)$

$\displaystyle x\equiv{y(mod a)}\Rightarrow{a}\mid{x-y}$

Therefore, $\displaystyle d_1\mid{a}$ and $\displaystyle a\mid{x-y}\Rightarrow{d_1\mid{x-y}}$

$\displaystyle d_1\mid{x-y}$ and $\displaystyle d_1\mid{x}\Rightarrow{d_1\mid{y}}$

$\displaystyle d_1$ is a common divisor of y and a

Therefore, $\displaystyle d_1\leq{d_2}$--------(1)

Similarly, $\displaystyle d_2$ is a common divisor of x and a

Therefore, $\displaystyle d_2\leq{d_1}$---------(2)

From (1) and (2),

$\displaystyle d_1=d_2\Rightarrow{(x,a)=(y,b)}$

Hope this helps. - Jan 30th 2010, 06:53 AMUnbeatable0
There's a short way to prove this.

$\displaystyle x\equiv y \pmod{a} \Leftrightarrow x=y+ka\:\:(k\in \mathbb{Z})$

Thus, using the fact that $\displaystyle \gcd(m,n)=\gcd(m+rn,n)$ for all $\displaystyle r\in \mathbb{Z}$, it follows that

$\displaystyle \gcd(x,a)=\gcd(y+ka,a)=\gcd(y,a)$ - Jan 31st 2010, 03:02 AMmatzerath
- Jan 31st 2010, 05:08 PMSudharaka
Dear matzerath,

Suppose, $\displaystyle (y+ka,a) = d_1$ and $\displaystyle (y,a) = d_2$ for $\displaystyle k\in{Z}$

Since, $\displaystyle (y+ka,a) = d_1$

$\displaystyle d_1\mid{y+ka}$ and $\displaystyle d_1\mid{a}$

Therefore, $\displaystyle d_1\mid{y}$ and $\displaystyle d_1\mid{a}$

$\displaystyle d_1\leq{d_2}$--------(1)

Also since, $\displaystyle d_2=(y,a)$

$\displaystyle d_2\mid{y}$ and $\displaystyle d_2\mid{a}$

Therefore, $\displaystyle d_2\mid{y+ka}$ and $\displaystyle d_2\mid{a}$

$\displaystyle d_2\leq{d_1}$---------(2)

By (1) and (2); $\displaystyle d_1=d_2\Rightarrow{(y+ka,a)=(y,a)}$

Hope this helps. - Feb 7th 2010, 05:05 AMmatzerath