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7-17-2008

Linear Diophantine Equations

A Diophantine equation is an equation which is to be solved over the integers.

A linear Diophantine equation of the form ax + by = c has solutions if and only if (a, b)

|c. There is a

similar result for linear Diophantine equations in more than 2 variables.

A Diophantine problem is one in which the solutions are required to be integers. Abusing terminology,Ill refer to Diophantine equations, meaning equations which are to be solved over the integers.

Example. x3 + y3 = z3 has many solutions over the reals; for example,

x = 1, , y = 1, z =3

2.

However, this equation has no nonzero integer solutions.

Example. Since (9, 100) = 1, there are integers x and y such that 9x + 100y = 1.For example, 9 (11) + 100 1 = 1, and 9 89 + 100 (8) = 1. That is, the Diophantine equation

9x + 100y = 1 has solutions in fact, infinitely many solutions.

Theorem. Let a,b,c Z. Consider the Diophantine equation

ax + by = c.

(a) If (a, b)| c, there are no solutions.(b) If (a, b) = d | c, there are infinitely many solutions of the form

x =b

dk + x0, y = a

dk + y0.

Here (x0, y0) is a particular solution, and k Z.Before I give the proof, Ill give some examples, and also discuss the three variable equation ax+by+cz =

d.

Example. Solve 6x + 9y = 21.

Since (6, 9) = 3 | 21, there are infinitely many solutions. By trial and error, x = 7, y = 7, is aparticular solution. Hence, the general solution is

x = 3k 7, y = 2k + 7.

For example, setting k = 5 produces the solution x = 8, y = 3.

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Example. Solve 6x + 9y = 5.

Since (6, 9) = 3| 5, the equation has no solutions.

Example. Solve

3x + 3y + 5z = 10.First, Ill factor (3, 3) out of the first two coefficients:

(3, 3)

3

(3, 3)x +

3

(3, 3)y

+ 5z = 10.

Notice that (3, 3) = 3, so those two fractions are actually integers. Im not simplifying3

(3, 3)so that

you can see whats going on.Let

w =3

(3, 3)x +

3

(3, 3)y.

The equation becomes

(3, 3)w + 5z = 10, or 3w + 5z = 10.

(3, 5) = 1 | 10, so this two variable equation is solvable. w = 5, y = 1, is a particular solution, so thegeneral solution is

w = 5s + 5, z = 3s 1.Now I have to find x and y:

w =3

(3, 3)x +

3

(3, 3)y, so w = x + y.

Thus,x + y = 5s + 5.

This is a two variable equation. Since (1, 1) = 1|

5s + 5, its solvable. x = 5, y = 5s, is a particularsolution. Therefore, the general solution is

x = t + 5, y = 5s t.

All together, the general solution to the original three variable equation is

x = t + 5, y = 5s t, z = 3s 1.

In general, if there is a solution to the linear Diophantine equation

a1x1 +

anxn = c,

the solution will depend on n 1 parameters exactly as youd expect from linear algebra.Proof. (two variable case) Consider the linear Diophantine equation

ax + by = c.

Case 1. Suppose (a, b)| c. If x and y solve the equation, then

(a, b) | ax + by = c.

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This contradiction shows that there cannot be a solution.

Case 2. Suppose (a, b) | c. Write c = k(a, b) for k Z. There are integers m and n such that

am + bn = (a, b).

Thenamk + bnk = (a, b)k = c.

Hence, x = km, y = kn, is a solution.Suppose x = x0, y = y0, is a particular solution. Then

a

b

dk + x0

+ b

a

dk + y0

=

ab

dk ab

dk + (ax0 + by0) = 0 + c = c.

This proves that x =b

dk + x0, y = a

dk + y0 is a solution for every k Z.

Finally, I want to show that every solution has this form. Suppose then that (x, y) is a solution. Thenax + by = c and ax0 + by0 = c imply

a(x x0) + b(y y0) = c c = 0.Therefore,

a

(a, b)(x x0) + b

(a, b)(y y0) = 0,

a

(a, b)(x x0) = b

(a, b)(y y0).

Nowa

(a, b)divides the left side, so it divides the right side. However,

a

(a, b),

b

(a, b)

= 1. Therefore,

a

(a, b) y y0, or y y0 = k a

(a, b)for some k.

Thus,

y = y0 + k a(a, b)

.

Substitute this back into the last x-y equation:

a

(a, b)(x x0) = b

(a, b)(y y0) = b

(a, b)k a

(a, b),

x x0 = k b(a, b)

,

x = x0

k

b

(a, b).

This is the result stated in the theorem (with an unimportant switch of k and k.)

c2008 by Bruce Ikenaga 3