Trying to prove $$\{14a + 29b : a, b \in \mathbb{Z}\} = \mathbb{Z}$$ The statement $\{14a + 29b : a, b \in \mathbb{Z}\} \subseteq \mathbb{Z}$ is trivial, but the other way eludes me. Any hints on how to think about this question? How to get started?
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Do you know Bezout's Lemma for the gcd? If not, can you see how to get $1$ as a linear combination? – Bill Dubuque Oct 09 '18 at 22:46
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4Let $a = -2n$, $b=n$. Then, we know that $$29\cdot n + 14\cdot(-2n) = 29n-28n=n$$ – Rushabh Mehta Oct 09 '18 at 22:47
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@BillDubuque I do not. But even if I did, I'd rather use elementary methods. – Jacopo Stifani Oct 09 '18 at 22:48
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"Bezout's Lemma" is an elementary method @JacopoStifani – Angina Seng Oct 10 '18 at 17:44
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Hint $\ 14a + (2\cdot 14\! +\! 1)\,b\, =\, 14(a\!+\!2b)+b.\,$ We can make it $\color{#c00}{=1} $ for $\,\color{#c00}{b=1}\,$ and $ \overbrace{a\!+\!2b = 0}^{\large a\ =\ -2b\ =\ -2}$
Hence $\,14(-2)+29(1) =\color{#c00} 1,\,$ and scaling this by $\,n\,$ shows how to obtain any integer that way.
Remark $\ $ The same works for $\,n\cdot a + (kn\!+\!1) b,\ $ i.e. $\ n(-k) + (kn\!+\!1)\cdot 1\, =\, 1$
It's the special case of the extended Euclidean algorithm for the GCD where it terminates in $1$ step, i.e. $\ \gcd(14,29) = \gcd(14,\color{#c00}1) = \color{#c00}1\ $ by $\ 29\bmod 14 \,=\, \color{#c00}1$
Bill Dubuque
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