CSE 311 Lecture 15:
Modular Exponentiation and Induction

Topics

Modular equations

A quick review of Lecture 14.

Modular exponentiation

A fast algorithm for computing $a^k\ \text{mod}\ m$.

Mathematical induction

A method for proving statements about all natural numbers.

Using induction

Using induction in formal and English proofs.

Example proofs by induction

Example proofs about sums and divisibility.

Bézout’s theorem and multiplicative inverses

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 compute $s$ and $t$ using the extended Euclidean algorithm.

If $\gcd{a}{m} = 1$, then $\mod{s}{m}$ is the multiplicative inverse of $a$ modulo $m$:

• $1 = \mod{(sa+tm)}{m} = \mod{sa}{m}$, so we have
• $\congruent{sa}{1}{m}$.

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

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

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

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

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.

The modular exponentiation problem: $\mod{a^k}{m}$

How would you compute $\mod{78365^{81453}}{104729}$?

To keep the intermediate results small, we use fast modular exponentiation.

Repeated squaring: $\mod{a^k}{m}$ for $k = 2^i$

If $k = 2^i$, we can compute $\mod{a^k}{m}$ in just $i$ steps.

Note that $\congruent{\mod{a}{m}}{a}{m}$ and $\congruent{\mod{b}{m}}{b}{m}$. So, we have $\mod{ab}{m} = \mod{((\mod{a}{m})(\mod{b}{m}))}{m}$.

Fast exponentiation: $\mod{a^k}{m}$ for all $k$

Note that 81453 is 10011111000101101 in binary.

$81453 = 2^{16} + 2^{13} + 2^{12} + 2^{11} + 2^{10} + 2^{9} + 2^{5} + 2^{3} + 2^{2} + 2^{0}$

$a^{81453} = a^{2^{16}} * a^{2^{13}} * a^{2^{12}} * a^{2^{11}} * a^{2^{10}} * a^{2^{9}} * a^{2^{5}} * a^{2^{3}} * a^{2^{2}} * a^{2^{0}}$

The fast exponentiation algorithm

// Assumes a > 0, k >= 0, m > 0.
public static long fastModExp(long a, long k, long m) {
    if (k == 0) {            // k = 0
        return 1;
    } else if (k % 2 == 0) { // k is even
        long tmp = fastModExp(a, k/2, m);
        return (tmp * tmp) % m;
    } else {                // k is odd
        long tmp = fastModExp(a, k-1, m);
        return (a * tmp) % m;
    }
}

$\mod{78365^{81453}}{104729} = 45235$

Using fast modular exponentiation: RSA encryption

Alice chooses random 512-bit (or 1024-bit) primes $p, q$ and exponent $e$.

Alice computes $m=pq$ and broadcasts $(m, e)$, which is her public key.

She also computes the multiplicative inverse $d$ of $\mod{e}{(p-1)(q-1)}$, which serves as her private key.

To encrypt a message $a$ with Alice’s public key, Bob computes $C = \mod{a^e}{m}$.

This computation uses fast modular exponentiation.

Bob sends the ciphertext $C$ to Alice.

To decrypt $C$, Alice computes $\mod{C^d}{m}$.

This computation also uses fast modular exponentiation.

It works because $\mod{C^d}{m} = a$ for $0 < a < m$ unless $p\vert a$ or $q \vert a$.

How would you prove this theorem?

Mods and exponents

For all integers $a, b, m > 0$ and $k \geq 0$, $\congruent{a}{b}{m} \rightarrow \congruent{a^k}{b^k}{m}$.

We don’t have a proof rule to say “perform this step repeatedly.”

Perform a step repeatedly with induction!

Induction is a logical rule of inference that applies (only) over $\N$.

If we know that a property $P$ holds for 0, and

we know that $\forall k. P(k)\rightarrow P(k+1)$, then

we can conclude that $P$ holds for all natural numbers.

// f(x) = x for all x >= 0.
public int f(int x) {
  if (x == 0) { return 0; }
  else        { return f(x - 1) + 1; }
}

Induction: how does it work?

Suppose that we are given $P(0)$ and $\forall k. P(k)\rightarrow P(k+1)$.

How does that give us $P(k)$ for a concrete $k$ such as $5$?

Using the induction rule in a formal proof

Using the induction rule in a formal proof: key parts

Translating to an English proof: the template

What is $\sum_{i=0}^{n}2^i$ for an arbitrary $n\in\N$?

Recall that $\sum_{i=0}^{n}2^i = 2^0 + 2^1 + \ldots + 2^n$.

Prove $\sum_{i=0}^{n}2^i = 2^{n+1}-1$ for all $n\in\N$

Prove $\sum_{i=0}^{n}i = n(n+1)/2$ for all $n\in\N$

What number divides $2^{2n}-1$ for every $n\in\N$?

Prove $3 \vert (2^{2n}-1)$ for all $n\in\N$