In my past exam I was asked to explain what Bezout coefficients are and what we can use them for. But I can't find any definitions online that answer the question specifically.
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1See https://en.wikipedia.org/wiki/Bézout%27s_identity. – lhf Aug 01 '17 at 13:12
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3Google "bezout" and look at the Wikipedia article. If this is not what you are looking for, you might want to give a little bit more context for us to find out what you are talking about. – Dirk Aug 01 '17 at 13:12
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The Bezout GCD identity is $\ \gcd(a,b) = a x + b y\ $ for some integers $\,x,y,\,$ i.e. the gcd can be written as an integral linear combination of its arguments. Here $\,x,y\,$ are the Bezout coefficients.
One common application of Bezout coefficients is computing modular inverses
$$ \gcd(a,b)= 1\,\Rightarrow\, ax+by = 1\,\Rightarrow\,\bmod b\!:\,\ ax\equiv 1\,\Rightarrow\, a^{-1}\equiv x$$
Here are some other interesting related applications.
Hermite's method of integrating rational functions.
Factoring $\ ab + x\ $ into power series when $\,\gcd(a,b) = 1$
By Bezout's GCD identity we may write $\ \color{#c00}{\bf 1} = ad\!-\!bc\,$
Recall the Catalan series $\rm\, C(x)\in \mathbb Z[[x]]\,$ satisfies $\,\color{#090}{C(x) - x\:C(x)^2\! = 1},\,$ so with $\,e = cd$
$\ \ \ (a - cx\:C(ex))\:(b+dt\:C(ex))\ =\ ab + \color{#c00}{\bf 1}\cdot x\:(\color{#090}{C(ex) - et\:C(ex)^2})\: =\ ab + x$
Bill Dubuque
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