Diophantine Equations
November 24, 2016
Euclid’s Lemma
Definition. An integer is said to be relatively prime to another integer b, if (a, b) = 1.
The following result is called Euclid’s Lemma. We will need it in the discussion which follows.
Lemma. (Euclid’s Lemma)
Let a, b and c be positive integers.
If a | bc, and a is relatively prime to b, i.e. (a, b) = 1, then a | c.
Proof. Assume a | bc, and (a, b) = 1. Since (a, b) = 1, there are integers x and y such that ax + by = 1.
Multiplying both sides of this equation by c, we have acx + bcy = c. Now clearly a | a and by assumption
a | bc. Thus, by previous result, a divides acx + bcy, because this is a linear combination of a and bc . Hence
a | c. Result follows.
Let us consider the general type of equation for which we want integer solutions. Such an equation is
called a Diophantine equation, named after the Greek mathematician, Diophantus, who was most likely of
Egyptian origin. (Diophantus is not the first to consider such research, the first in recorded history was the
Indian mathematician Brahmagupta.)
Definition. A Diophantine equation is an equation in which all the coefficients are integers and in which
we are interested only in integer solutions.
Any equation with integer coefficients may not be considered as a Diophantine equation, depending on
our point of view.
For example, the (linear) equation 3x + 5y = 1 has an infinite number of solutions in general (x = 13 , y =
0 or x = 1, y = − 25 and so on). It also has integer solutions (such as x = 2, y = −1; or x = −3, y = 2);
therefore as a Diophantine equation, it has solutions. On the other hand, the (linear) equation 4x + 6y = 1
also has an infinite number of solutions in general (for example x = 14 , y = 0;and x = 1, y = − 12 ). However,
we will soon prove that it has no integer solutions; so, as a Diophantine equation, it has no solutions.
As another example, consider the equation x3 + y 3 = z 3 . This also has many solutions in general, such as
√
x = y = 1, z = 3 2; however, if we consider this as a Diophantine equation - that is, if we are only interested
in integral solutions - then it is known there are no solutions other than x = 0, y = z, or x = 0, x = z, or
z = 0, y = −x.
There are no general techniques that may be used to solve all Diophantine equations. However, there is
one kind of Diophantine equation that can always be solved - a linear Diophantine equation. In this course,
we investigate a method for attacking linear Diophantine equations.
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Linear Diophantine Equations
Definition. A linear Diophantine equation in two unknowns is an equation of the form
ax + by = c,
where a, b and c are known integers,where we are interested only in integer solutions - integers x and y which
satisfy the equation.
Similarly, a linear Diophantine equation in three unknowns is of the form
ax + by + cz = d,
etc.
In this course we restrict our attention to linear Diophantine equations in two unknowns.
We now develop the theory for solving such equations.
Theorem 1. Let a and b be integers with d = (a, b). The equation ax + by = c has no integral solutions if
d - c. If d | c, then there are infiniteely many integral solutions. Moreover, if x = x0 , y = y0 is a particular
solution of the equation, then all solutions are given by
x = x0 +
b
n,
d
y = y0 −
a
d
n,
where n is an integer.
Proof. Omitted.
Example. By the above theorem, there are no integer solutions of the diophantine equation 2x + 4y = 1
because (2, 4) = 2 but 2 - 1.
Example. By the above theorem, there are no integer solutions of the diophantine equation 15x + 6y = 7
because (15, 6) = 3 but 3 - 7.
Example. Determine how many integer solutions there are for the equation 21x + 14y = 70.
Solution:
There are many solutions for the given equation, because (21, 14) = 7 and 7 | 70. To find these solutions
we use the Euclidean algorithm to establish the result (21, 14)=7.
21 = 14 · 1 + 7 .....(1)
14 = 7 · 2 .............(2)
Thus (21, 14) = 7 .
But by equation (1), we have: 7 = 21 · 1 + 14 · (−1), so that
21 · (10) + 14 · (−10) = 70
Hence, x0 = 10, y0 = −10 is a particular solution. Therefore, all solutions are given by
x = 10 +
14
21
n, y = −10 − n
7
7
where n is any integer. i.e.
x = 10 + 2n, y = −10 − 3n, n ∈ Z.
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