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Let $a,b,c\in\mathbb Z^+$. If $a\equiv b\pmod c$ then $\gcd(a,c)=\gcd(b,c)$. Use this to show that there are no $ x, y $ such that $ x + y = 100 $ and $ (x, y) = 3 $.

Since $a\equiv b\pmod c$, then $a=b+ck$ with $k\in\mathbb Z$. Let $d=\gcd(a,c)$, then exists $x,y\in\Bbb Z^+$ such that $$\begin{align*} d&=ax+cy\\ &=(b+ck)x+cy\\ &=bx+ckx+cy\\ &=bx+c(kx+y)\\ &=bx+cz. \end{align*}$$ This shows that $ d = \gcd(b, c) $. Is my proofcorrect? And how do I use this for the particular result. Thanks a lot.

Bill Dubuque
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Andrew Madison
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    This shows only that $\gcd(b, c) \mid d$. It doesn't show equality. You need to separately show $d \mid b$ (you already have $d \mid c$). – Robert Shore Aug 05 '21 at 23:56
  • Doing $ p = \gcd (b, c) $ and reasoning in the same way, I get to the other one, right? and would have equality? – Andrew Madison Aug 05 '21 at 23:57
  • Immediate consequence of the theorem (proved in the linked dupe), viz. $, \bmod x!:\ \color{#c00}{y} = 100!-!x\equiv \color{#c00}{100},$ so by the Theorem we conclude that $,(x,\color{#c00}y) = (x,\color{#c00}{100})\mid 100,$ so $,(x,y)\neq 3\ \ $ – Bill Dubuque Aug 06 '21 at 01:46

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A slight rephrase:

Let $d$ be any common divisor of $a$ and $c$, then consider $b = a-ck$ with $k\in\mathbb Z$, $d$ is a divisor of the right hand side, so $d\mid b$.

i.e. All common divisors of $a$ and $c$ are also divisors of $b$, and hence are common divisors of $b$ and $c$.

Similarly, by considering $a=b+ck$, all common divisors of $b$ and $c$ are also common divisors of $a$ and $c$.

So the set of common divisors of $a$ and $c$ is the same as the set of common divisors of $b$ and $c$. The greatest elements of the two sets are also the same.

peterwhy
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