Bezout's Lemma

Given integers x x and y y, a linear combination of x x and y y has the form ax+by ax + by. Bezout's Lemma relates the question of finding the greatest common divisor of xx and yy to the question of finding linear combinations of xx and yy.

Bezout's Lemma: Given integers x x and y y, there exists integers a a and b b such that ax+by=gcd(x,y) ax+by = \gcd(x,y) .

Technique

Applying the Euclidean Algorithm backwards gives an algorithm to obtain the integer values of a a and b b. We show how to obtain integers a a and b b such that 16457a+1638b=7 16457a + 1638b = 7. These numbers are calculated in Euclidean Algorithm.

7=2114×1=21×114×1=21(7721×3)×1=77×1+21×4=77×1+(163877×21)×4=1638×477×85=1638×4(164571638×10)×85=16457×(85)+1638×854\begin{array} {llllllllll} 7 & = & 21 &- & 14 \times 1 \\ & = & 21 \times 1 & - & 14 \times 1\\ & = & 21 &- & (77 - 21 \times 3) \times 1\\ & = & - 77 \times 1 &+& 21 \times 4\\ & = & - 77 \times 1 &+ & (1638 - 77 \times 21) \times 4 \\ &= &1638 \times 4 &-& 77 \times 85\\ & = & 1638 \times 4 &- & (16457 - 1638 \times 10) \times 85 \\ & = &16457 \times (-85) &+& 1638 \times 854\\ \end{array}

Worked Examples

1. Given integers x,y x, y, describe the set of all integers N N that can be expressed in the form N=ax+by N=ax+by, where a,b a, b are integers.

Solution: Let k=gcd(x,y) k = \gcd(x,y). Then any integer of the form kn kn, where n n is an integer, can be expressed as ax+by ax+by. We already know that this condition is a necessary condition, so to show that it is sufficient, Bezout's Lemma tells us that there exists integers a,b a', b' such that k=ax+by k = a' x + b'y. Therefore,

kn=(an)x+(bn)y. kn = (a'n) x + (b'n) y.

 

2. [Modulo Arithmetic Property I - Multiplicative inverses] Show that if a,b a, b are integers such that gcd(a,n)=1 \gcd(a,n)=1, then there exists an integer x x such that ax1(modn) ax \equiv 1 \pmod{n}.

Solution: Since gcd(a,n)=1 \gcd(a,n)=1, Bezout's Lemma implies there exists integers x,y x, y such that ax+ny=gcd(a,n)=1 ax + n y = \gcd (a,n) = 1. Then

1ax+nyax(modn). 1 \equiv ax+ny \equiv ax \pmod{n} .

Note by Calvin Lin
5 years, 7 months ago

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I learned a lot! Thanks :D

Astro Enthusiast - 5 years, 4 months ago

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I did not understand what the 1st worked example mean. Any number of form kn can be represented in the form ax+by but that does not mean that all nos of form ax+by will be of form kn. Actually just explain what the question demands?

Chandrachur Banerjee - 5 years, 3 months ago

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The question asks you to classify all numbers that can be written in the form ax+by ax + by for given integers xx and yy. The claim is that these numbers are exactly those of the form kn k n , wherek=gcd(x,y) k = \gcd (x,y) and nn is any integer.

As you realized, there are 2 parts to this.
First, we have to show that any number of the form ax+byax + by can be written as knkn for some nn. This is obvious because we can let x=kx,y=ky x = k x^*, y = ky^* then ax+by=akx+bky=k(ax+by) ax + by = a k x^* + bky^* = k ( ax^* + by^* ) .
Second, we have to show that any number of the form kn kn can be written as ax+by ax + by for some aa and bb. This follows by applying Bezout's lemma, which tells us that there exists integers which satisfy k=ax+by k = a' x + b' y , and hence kn=(an)x+(bn)y kn = ( a'n) x + (b'n) y .

Calvin Lin Staff - 5 years, 3 months ago

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thanks....

Shatrughna Kumar - 5 years, 3 months ago

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