How would I go by proving this statement?
What I did was I tried using Proposition GCD Of One, so that
$(11a+5)x + (2a+1)y = 1$, and
$(11x+2y)a + (5x+y) = 1$.
But I have no idea what to do from here. What can I do?
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Hint: Can you justify the following?
\begin{align*} \gcd(11a + 5, 2a + 1) &= \gcd\Big((11a + 5) - 5(2a + 1), 2a + 1\Big) \\ &= \gcd(a, 2a + 1) \end{align*}
Can you finish it from here?
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Sorry, but could you explain how the first line works? – Haxify Apr 15 '14 at 00:33
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@Haxify $$\gcd(a, c) = \gcd(a - c, c)$$ – Apr 15 '14 at 00:35
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Oh, okay I see how that works. Thank you! – Haxify Apr 15 '14 at 00:37
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Then is it correct to say that $\gcd(11a+5,2a+1)=\gcd(a,2a+1)=\gcd(1,a)$? – Haxify Apr 15 '14 at 00:47
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@Haxify Exactly. – Apr 15 '14 at 00:47
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Great. Thank you! – Haxify Apr 15 '14 at 00:49
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Hint $\ $ If $\,d\,$ is a common factor then $\,{\rm mod}\ d\!:\ 2a+1\equiv 0\equiv 11a+5.\,$ Now eliminate $\,a\,$ by cross multiplying $\ 0\equiv \color{#0a0}{11}(2a+1)-\color{#c00}2(11a+5) \equiv \color{#0a0}{11}-\color{#c00}{10}\equiv 1\ $ i.e. $\,0\equiv 1,\,$ so $\, d\mid 1\!-\!0,\,$ so $\,d = 1.$
Remark $\ $ More conceptually, put the fractions for $\,-a\,$ over the common denominator $\,22$
$$ {\rm mod}\ d\!:\ \ \frac{1}{2} \equiv -a\,\equiv \frac{5}{11}\ \Rightarrow\ \frac{\color{#0a0}{11}}{22}\,\equiv\, \frac{\color{#c00}{10}}{22}\qquad$$
Note $ $ The fractions uniquely exist mod $\,d\,$ since $\,2,11\,$ are invertible, being coprime to $\,d\,$ (else $2\mid d\mid 2a+1\,\Rightarrow 2\mid 1,\,$ and $\,11\mid d\mid 11a+5\,\Rightarrow\, 11\mid 5\,)$.
Bill Dubuque
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