Solving Linear Diophantine Equations: Finding Integral Solutions, Assignments of Number Theory

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Linear Diophantine Equations

Ryan C. Daileda

Trinity University

Number Theory

Introduction

A Diophantine equation is any equation (usually polynomial) in one or more variables that is to be solved in Z.

For example, a pythagorean triple is a solution to the Diophantine equation x^2 + y 2 = z^2 ,

such as (3, 4 , 5) or (5, 12 , 13).

Solving Diophantine equations is substantially more difficult than solving equations over R, say, since Z is discrete while R is continuous and we can appeal to the tools of analysis.

Today we will consider the simplest of all Diophatine equations: linear Diophantine equations in two variables.

• The line 3x + 5y = 7 in R
• The lattice Z

We immediately observe (and can easily verify) the solutions:

x = −6 and y = 5; x = −1 and y = 2; x = 4 and y = −1; x = 9 and y = −4;

which are regularly spaced on the line.

On the other hand, the equation

4 x + 6y = 3

has no integral solutions (why not?).

That is, the line with equation 4x + 6y = 3 completely avoids the lattice Z^2.

The line 4x + 6y = 3.

This condition is also sufficient. To see this, suppose (a, b)|c.

Write c = (a, b)d with d ∈ Z.

By B´ezout’s lemma, we can find r , s ∈ Z so that

ar + bs = (a, b).

If we multiply both sides by d we obtain

a(rd) + b(sd) = (a, b)d = c.

Thus x = rd, y = sd is a solution to ax + by = c.

Summary

Theorem 1 Let a, b, c ∈ Z with a, b 6 = 0. The Diophantine equation

ax + by = c

has a solution if and only if (a, b)|c. In this case, one solution is given by x 0 = r

c (a, b)

, y 0 = s

c (a, b)

where r , s ∈ Z satisfy ar + bs = (a, b).

Because we can use the Euclidean algorithm to effectively compute (a, b), r and s, we can always produce at least one solution to ax + by = c (when it exists).

Solving ax + by = c

We begin by assuming (a, b) = 1.

Since 1|c for all c ∈ Z, we are guaranteed the existence of at least one (integral) solution x = x 0 , y = y 0.

Suppose x = x 1 , y = y 1 is another solution. We then have

ax 0 + by 0 = c, ax 1 + by 1 = c.

Subtraction yields

a(x 0 − x 1 ) + b(y 0 − y 1 ) = 0 ⇒ a(x 0 − x 1 ) = −b(y 0 − y 1 )

⇒ a|b(y 0 − y 1 ).

Since (a, b) = 1, Euclid’s lemma implies that a|(y 0 − y 1 ). Write

y 0 − y 1 = ak.

Then y 1 = y 0 − ak and back substitution yields

a(x 0 − x 1 ) = −abk ⇒ x 0 − x 1 = −bk ⇒ x 1 = x 0 + bk.

Thus, every solution to ax + by = c has the form

x = x 0 + bk, y = y 0 − ak,

for some k ∈ Z.

Examples

Example 2 (Continued) Finish solving 3x + 5y = 7.

Solution. It’s easy to note that 2 · 3 + (−1)5 = 1.

So we may take r = 2 and s = −1.

Thus the general solution is given by

x = 14 + 5k, y = − 7 − 3 k,

with k ∈ Z.

Example 3

Describe the set of solutions to the Diophantine equation 313 x + 510y = 2.

Solution. If we apply the Euclidean algorithm, it takes 8 divisions to determine that (313, 510) = 1.

The first 7 quotients are q = 1, 1 , 1 , 1 , 2 , 3 , 5, and multiplication of the associated matrices

1 −q

(in the opposite order) yields the matrix (^) ( ∗ ∗ 143 − 233

Solving ax + by = c in General

Now let a, b, c ∈ Z (with a, b 6 = 0) be arbitrary integers satisfying (a, b)|c. The Diophantine equation

ax + by = c

is then equivalent to the equation

a (a, b)

x +

b (a, b)

y =

c (a, b)

Since (^) ( a (a, b)

b (a, b)

the latter equation can be solved using our previous result.

We write

a (a, b)

r +

b (a, b)

s = 1 or ar + bs = (a, b)

and set

x = r

c (a, b)

b (a, b)

k,

y = s

c (a, b)

a (a, b)

k,

with k ∈ Z.

We’ve now completely solved the Diophantine ax + by = c.