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QUESTION

For each of the following diophantine equations

\begin{description}

\item[(i)]
decide whether or not the solution exists,

\item[(ii)]
if a solution exists, find the general solution,

\item[(iii)]
Find the solution in which $x$ takes the smallest possible
positive integer value.

\begin{description}

\item[(a)]
$15x+28y=5$

\item[(b)]
$18x+27y=7$

\item[(c)]
$12x+2y=8$

\end{description}

\end{description}



ANSWER

\begin{description}

\item[(a)]

\begin{description}

\item[(i)]
gcd(15,28)=1, which divides 5, so solutions exist.

\item[(ii)]
We seek a particular solution $(x_0,y_0)$. If we can find it, then
by th.1.11, the general solution is $(x_0+28t,y_0-15t)$ where
$t\in Z$.

For a particular solution, we first use the Euclidean algorithm to
solve $15x+28y=1$:

\begin{eqnarray*}
28&amp;=&amp;15.1+13\\ 15&amp;=&amp;13.1+2\\ 13&amp;=&amp;2.6+1\\ 2&amp;=&amp;2.1+0
\end{eqnarray*}

Thus
gcd$(15,28)=1=13-2.6=13-(15-13).6=13.7-15.6=(28-15).7-15.6=28.7-15.13$.
Multiplying by $5,\ 5=15-65+28.35$, so that $x_0=-65$ and $y_0=35$
is a solution.

Hence $x=-65+28t,\ y=35-15t\ (t\in Z)$ is the general solution.

\item[(iii)]
To make $x&gt;0$ we need $-65+28t&gt;0$, and the smallest $t\in Z$ for
which this happens (and hance gives the smallest positive $x$) is
3. Thus the relevant solution is $x=19,\ y=-10$.

\end{description}

\item[(b)]
gcd(18,27)=9, which does not divide 7, so this equation has no
solutions.

\item[(c)]

\begin{description}

\item[(i)]
gcd(12,2)=2, which divides 8, so solutions exist.

\item[(ii)]
If $(x_0,y_0)$ is a particular solution, then by th.1.11, the
general solution is $x=x_0+\left(\frac{2}{2}\right)t,\
y=y_0-\left(\frac{12}{2}\right)t$, i.e. $x=x_0+t,\ y=y_0-6t$,
where $t\in Z$.

To find a particular solution, we could follow the general policy,
as we did in (a), but this time a short cut is available: if
$12x+2y=8$ then, on dividing by 2, $6x+y=4$, and there is an
obvious solution with $x=0,\ y=4$. The general solution is,
therefore, $x=t,\ y=4-6t$.

\item[(iii)]
The solution where $x$ takes the smallest positive value is given
by $t=1$. It is $x=1,\ y=-2$.

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