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| − | '''Bezout's Lemma''' states that if two integers <math>x</math> and <math>y</math> satisfy <math>gcd(x,y)=1</math>, then there exist integers <math>\alpha</math> and <math>\beta</math> such that <math>x\alpha+y\beta=1</math>. In other words, there exists a linear combination of <math>x</math> and <math>y</math> equal to <math>1</math>.
| + | #REDIRECT[[Bézout's Identity]] |
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| − | ==Proof==
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| − | Since <math>gcd(x,y)=1</math>, <math>lcm(x,y)=xy</math>. So <math>\alpha=y</math> is the first time that <math>x\alpha\equiv 0\bmod{y}</math>, and it is there that the modular residues begin repeating. Now if for all integers <math>0<a,b<n</math>, we have that <math>xa\neq xb\bmod{y}</math>, then one of those <math>n-1</math> integers must be 1 from the [[Pigeonhole Principle]]. Assume for contradiction that <math>xa\equiv xb\bmod{y}</math>. Thus it repeats, and one of <math>a</math> or <math>b</math> must be <math>\geq n</math>, which is opposite of what we had. Thus there exists an <math>\alpha</math> such that <math>x\alpha\equiv 1\bmod{y}</math>, and the same proof holds for <math>\beta</math>.
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| − | Since <math>x\alpha +y\beta</math> is equivalent to 1 mod x and mod y, and <math>gcd(x,y)=1</math>, <math>x\alpha +y\beta \equiv 1\bmod{xy}</math>. Lets say that <math>x\alpha+y\beta=xy\gamma +1</math> for some integer <math>\gamma</math>. We can subtract <math>y\gamma</math> from <math>\alpha</math> and plug that in to get
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| − | <math>x(\alpha-y\gamma)+y\beta=xy\gamma+1-xy\gamma=1</math>.
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| − | Thus there does exist integers <math>\alpha</math> and <math>\beta</math> such that <math>x\alpha+y\beta=1</math>.
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| − | ==See also==
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| − | [[Category:Number theory]]
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| − | {{stub}}
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