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Bezout's Identity & Extended Euclidean Algorithm cheat sheet - grade 11-12

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Math Grade 11-12

Bezout's Identity & Extended Euclidean Algorithm Cheat Sheet

A printable reference covering Bézout coefficients, the greatest common divisor, the Extended Euclidean Algorithm, modular inverses, and linear Diophantine equations for grades 11-12.

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Bézout's Identity connects divisibility, greatest common divisors, and integer linear combinations. This cheat sheet helps students find integers that solve equations of the form ax+by=gcd(a,b)ax + by = \gcd(a,b). The Extended Euclidean Algorithm gives a reliable way to compute both the greatest common divisor and the needed coefficients.

These tools are especially important in number theory, modular arithmetic, and cryptography.

The core idea is that for any integers aa and bb, not both zero, there exist integers xx and yy such that ax+by=gcd(a,b)ax + by = \gcd(a,b). The Euclidean Algorithm finds gcd(a,b)\gcd(a,b) by repeated division with remainders. The extended version works backward through those divisions to express the gcd as a linear combination of the original numbers.

If gcd(a,m)=1\gcd(a,m)=1, Bézout's Identity also gives the modular inverse of aa modulo mm.

Key Facts

  • Bézout's Identity states that for integers aa and bb, not both zero, there exist integers xx and yy such that ax+by=gcd(a,b)ax + by = \gcd(a,b).
  • The Euclidean Algorithm uses repeated division: a=bq+ra = bq + r, where 0r<b0 \le r < |b|, then replaces (a,b)(a,b) with (b,r)(b,r).
  • The last nonzero remainder in the Euclidean Algorithm is gcd(a,b)\gcd(a,b).
  • The Extended Euclidean Algorithm rewrites remainders backward until gcd(a,b)\gcd(a,b) is expressed as ax+byax + by.
  • If gcd(a,b)=1\gcd(a,b)=1, then aa and bb are relatively prime and there exist integers xx and yy such that ax+by=1ax + by = 1.
  • A linear Diophantine equation ax+by=cax + by = c has integer solutions exactly when gcd(a,b)\gcd(a,b) divides cc.
  • If ax+by=gcd(a,b)ax + by = \gcd(a,b), then multiplying by cgcd(a,b)\frac{c}{\gcd(a,b)} gives one solution to ax+by=cax + by = c when gcd(a,b)c\gcd(a,b) \mid c.
  • If ax+my=1ax + my = 1, then xx is a modular inverse of aa modulo mm, so a1x(modm)a^{-1} \equiv x \pmod{m}.

Vocabulary

Greatest common divisor
The greatest common divisor gcd(a,b)\gcd(a,b) is the largest positive integer that divides both aa and bb.
Bézout coefficients
Bézout coefficients are integers xx and yy that satisfy ax+by=gcd(a,b)ax + by = \gcd(a,b).
Euclidean Algorithm
The Euclidean Algorithm is a repeated division process used to find gcd(a,b)\gcd(a,b) efficiently.
Extended Euclidean Algorithm
The Extended Euclidean Algorithm finds both gcd(a,b)\gcd(a,b) and integers xx and yy such that ax+by=gcd(a,b)ax + by = \gcd(a,b).
Relatively prime
Two integers aa and bb are relatively prime when gcd(a,b)=1\gcd(a,b)=1.
Modular inverse
A modular inverse of aa modulo mm is a number xx such that ax1(modm)ax \equiv 1 \pmod{m}.

Common Mistakes to Avoid

  • Stopping before the last nonzero remainder is a mistake because gcd(a,b)\gcd(a,b) is the last nonzero remainder, not the final remainder 00.
  • Using quotient signs incorrectly during back substitution is wrong because each equation must be rearranged exactly, such as r=abqr = a - bq.
  • Assuming Bézout coefficients are unique is wrong because if one pair (x,y)(x,y) works, infinitely many related pairs can also work.
  • Claiming every equation ax+by=cax + by = c has integer solutions is wrong because solutions exist only when gcd(a,b)c\gcd(a,b) \mid c.
  • Forgetting to reduce a negative modular inverse is a mistake because an answer like x=3x=-3 should often be written as xm3(modm)x \equiv m-3 \pmod{m}.

Practice Questions

  1. 1 Use the Extended Euclidean Algorithm to find integers xx and yy such that 252x+198y=gcd(252,198)252x + 198y = \gcd(252,198).
  2. 2 Find the modular inverse of 1717 modulo 4343 using Bézout's Identity.
  3. 3 Determine whether the equation 84x+30y=1884x + 30y = 18 has integer solutions, and if it does, find one solution.
  4. 4 Explain why aa has a modular inverse modulo mm exactly when gcd(a,m)=1\gcd(a,m)=1.