6. Linear Diophantine equations
We consider the problem of trying to find integral solutions to linear
equations with integer coefficients. The general problem of finding
integral solutions to polynomial equations with integer coefficients is
called a Diophantine problem, so we are looking at linear Diophantine
equations.
The general linear Diophantine equation in two variables has the
form
ax +by =c,
where a,band c∈Z. Our goal is to describe all solutions (x, y) with
integer coordinates, x∈Zand y∈Z.
Geometrically the equation ax +by =cdescribes a line in the plane
R2. It is possible that this line misses the integral points Z2. However
if there is one integral point on the line then there are infinitely many.
We first start by solving a special case, when c= 0 and (a, b) = 1.
When c= 0 one solution is clear x=y= 0, the origin (0,0) is a point
of the line
ax +by = 0.
On the hand we could increase xby band decrease yby a, to get
another point (b, −a) of the line. More generally if k∈Zis an integer
then could we could increase xby kb and decrease yby ka, to get
another point (kb, −ak) of the line.
On the other hand, if you consider the equation
ax =−by,
it is clear that every point of the line has the form (bt, −at), for some
real number t. For this to be an integer point, t=k/l has to be
rational, where kand l > 0 are coprime integers and we must have
bk/l ∈Zand ak/l ∈Z. Suppose that l6= 1. Then some prime p
divides l.pdoes not divide k, as (k, l ) = 1 and pdoes not divide both
aand b, as (a, b) = 1, a contradiction. Thus t=kand the general
solution to the equation
ax +by = 0
is (kb, −ka), where k∈Zis an integer.
Now we turn to the case when (a, b) = 1 but cis arbitrary. In this
case we may find (λ, µ) such that λa +µb = 1. If we multiply this
equation by cthen we get
a(cλ) + b(cµ) = c.
Hence, if we put
x0=cλ and y0=cµ
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