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| − | '''Bezout's Lemma''' states that if <math>x</math> and <math>y</math> are nonzero integers and <math>g = \gcd(x,y)</math>, then there exist integers <math>\alpha</math> and <math>\beta</math> such that <math>x\alpha+y\beta=g</math>. In other words, there exists a linear combination of <math>x</math> and <math>y</math> equal to <math>g</math>.
| + | #REDIRECT[[Bézout's Identity]] |
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| − | In particular, if <math>x</math> and <math>y</math> are [[relatively prime]] then there are integers <math>\alpha</math> and <math>\beta</math> for which <math>x\alpha+y\beta=1</math>.
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| − | ==Proof==
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| − | Let <math>x = gx_1</math>, <math>y = gy_1</math>, and notice that <math>\gcd(x_1,y_1) = 1</math>.
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| − | Since <math>\gcd(x_1,y_1)=1</math>, <math>\text{lcm}(x_1,y_1)=x_1y_1</math>. So <math>\alpha=y_1</math> is smallest positive <math>\alpha</math> for which <math>x\alpha\equiv 0\pmod{y}</math>. Now if for all integers <math>0\le a,b<y_1</math>, we have that <math>x_1a\not\equiv x_1b\pmod{y_1}</math>, then one of those <math>y_1</math> integers must be 1 from the [[Pigeonhole Principle]]. Assume for contradiction that <math>x_1a\equiv x_1b\pmod{y_1}</math>, and WLOG let <math>b>a</math>. Then, <math>x_1(b-a)\equiv 0\pmod y_1</math>, and so as we saw above this means <math>b-a\ge y_1</math> but this is impossible since <math>0\le a,b<y_1</math>. Thus there exists an <math>\alpha</math> such that <math>x_1\alpha\equiv 1\pmod{y_1}</math>.
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| − | Therefore <math>y_1|(x_1\alpha-1)</math>, and so there exists an integer <math>\beta</math> such that <math>x_1\alpha - 1 = y_1\beta</math>, and so <math>x_1\alpha + y_1\beta = 1</math>. Now multiplying through by <math>g</math> gives, <math>gx_1\alpha + gy_1\alpha = g</math>, or <math>x\alpha+y\beta = g</math>.
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| − | Thus there does exist integers <math>\alpha</math> and <math>\beta</math> such that <math>x\alpha+y\beta=g</math>.
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| − | ==See also==
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| − | [[Category:Number theory]]
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| − | {{stub}}
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