CSE 311 Lecture 14: Euclidean Algorithm and Modular Equations

Topics

Primes and GCD

A quick review of Lecture 13.

Extended Euclidean algorithm

Bézout’s theorem and the extended Euclidean algorithm.

Modular equations

Solving modular equations with the extended Euclidean algorithm.

Primes and composites: definitions and theorems

Greatest common divisor (GCD): definition

Greatest common divisor (GCD)

The greatest common divisor of integers $a$ and $b$, written as $\gcd{a}{b}$, is the largest integer $d$ such that $d\vert a$ and $d\vert b$.

We can compute GCDs efficiently using the Euclidean algorithm. Invented in 300 BC!

Euclidean algorithm: review

Euclidean algorithm is based on two useful facts:

$\gcd{a}{0} = a$ for all positive integers $a$.

$\gcd{a}{b} = \gcd{b}{\mod{a}{b}}$ for all positive integers $a$ and $b$.

// Assumes a >= b >= 0.
public static int gcd(int a, int b) {
  if (b == 0)
    return a;             // GCD(a, 0) = a
  else  
    return gcd(b, a % b); // GCD(a, b) = GCD(b, a mod b)
}

Bézout’s theorem about GCDs

Bézout’s theorem

If $a$ and $b$ are positive integers, then there exist integers $s$ and $t$ such that $\gcd{a}{b} = sa + tb$.

We can extend Euclidean algorithm to find $s$ and $t$ in addition to computing $\gcd{a}{b}$.

Extended Euclidean algorithm

Extended Euclidean algorithm

Extended Euclidean algorithm

Multiplicative inverse $\mod{}{m}$

Suppose $\gcd{a}{m} = 1$.

By Bézout’s theorem, there exist integers $s$ and $t$ such that $sa + tm = 1$.

$\mod{s}{m}$ is the multiplicative inverse of $a$ modulo $m$: $\congruent{(\mod{s}{m})a}{1}{m}$

To see why, note that $\congruent{sa}{1}{m}$ and $\congruent{s}{\mod{s}{m}}{m}$, so by the multiplication property, $\congruent{(\mod{s}{m})a}{sa}{m}$, and by transitivity of congruence modulo $m$, we have that $\congruent{(\mod{s}{m})a}{1}{m}$.

So, we can compute multiplicative inverses with the extended Euclidean algorithm. These inverses let us solve modular equations.

Using multiplicative inverses to solve modular equations

Solve: $\congruent{7x}{1}{26}$

Solving a more general equation

Solve: $\congruent{7y}{3}{26}$

We computed that 15 is the multiplicative inverse of 7 modulo 26:

That is, $\congruent{7 * 15}{1}{26}$.

By the multiplication property of mod, we have

$\congruent{7 * 15 * 3}{1 * 3}{26}$.

So, any $\congruent{y}{15 * 3}{26}$ is a solution.

That is, $y = 19 + 26k$ for any $k\in\Z$ is a solution.

Solving equations modulo a prime number

$\gcd{a}{m}=1$ if $m$ is prime and $0 < a < m$, so we can always solve modular equations for prime $m$.

A useful proof technique based on modular equations

Suppose that $x,y\in\Z$ and $(x, y)$ satisfies linear equations

$ax + by = c$ and $dx + ey = f$,

where $a,b,c,d,e,f$ are integer coefficients.

Then $(x, y)$ also satisfies the corresponding equations mod $m > 0\in\Z$:

$\congruent{ax + by}{c}{m}$ and $\congruent{dx + ey}{f}{m}$.

The reverse doesn’t hold. Can you think of a counterexample?

$(0, 0)$ is a solution to $\congruent{x + y}{2}{2}$ and $\congruent{2x + 2y}{4}{2}$.

But it’s not a solution to $x + y = 2$ and $2x + 2y = 4$.

The contrapositive is a useful proof technique:

You can prove that a system of linear equations with integer coefficients has no integer solutions by showing that those equations modulo $m$ have no solutions.