How can we solve equations like $13x+29y=42$ or $2x+4y=13$ with
the solutions $x$ and $y$ being integers? Equations with integer
solutions are called Diophantine equations after Diophantus who
lived about 250 AD but the methods described here go back to
Euclid (about 300 BC) and earlier. When people hear the name
Euclid they think of geometry but the algorithm described here
appeared as Proposition 2 in Euclid's Book 7 on Number Theory.
First we notice that $13x+29y=42$ has many solutions, for example
$x=1$, $y=1$ and $x=30$, $y=-12$. Can you find others (it has
infinitely many solutions)? We also notice that $2x+4y=13$ has no
solutions because $2x+4y$ must be even and $13$ is odd. Can you
find another equation that has no solutions?
If we can solve $3x+5y=1$ then we can also solve $3x+5y=456$. For
example, $x=2$ and $y=-1$ is a solution of the first equation, so
that $x=2\times 456$ and $y=-1\times 456$ is a solution of the
second equation. The same argument works if we replace $456$ by
any other number, so that we only have to consider equations with
$1$ on the right hand side, for example $P x+Q y=1$. However if
$P$ and $Q$ have a common factor $S$ then $P x+Q y$ must be a
multiple of $S$ so we cannot have a solution of $P x+Q y=1$
unless $S=1$. This means that we should start by considering
equations $P x+Q y=1$ where $P$ and $Q$ have no common
factor.
Let us consider the example $83x+19y=1$. There is a standard
method, called Euclid's Algorithm, for solving such equations. It
involves taking the pair of numbers $P=83$ and $Q=19$ and
replacing them successively by other pairs $(P_k,Q_k)$. We
illustrate this by representing each pair of integers $(P_k,Q_k)$
by a rectangle with sides of length $P_k$ and $Q_k$.
Draw an $83$ by $19$ rectangle and mark off $4$ squares of side
$19$, leaving a $19$ by $7$ rectangle.
This diagram represents the fact that
$83=4\times 19+7$
In a few steps we shall split this rectangle into `compartments'
to illustrate the whole procedure for solving this equation. (You
may like to try the java applet
Solving
with Euclid's Algorithm which draws the rectangles and
carries out all the steps automatically to solve equations of the
form $P x+Q y=1$). We repeat this process using the $19$ by $7$
rectangle to obtain two squares of side $7$, and a $7$ by $5$
rectangle. Next, the $7$ by $5$ rectangle splits into a square of
side $5$, and a $5$ by $2$ rectangle.
The $5$ by $2$ rectangle splits into two squares of side $2$, and
a $2$ by $1$ rectangle. The $2$ by $1$ rectangle splits into two
squares of side $1$ with nothing left over and the process
finishes here as there is no residual rectangle. These diagrams
illustrate the following equations:
\begin{eqnarray} 83 & = & 4\times 19 & + & 7\\
19 & = & 2\times 7 & + & 5\\ 7 & = &
1\times 5 & + & 2\\ 5 & = & 2\times 2 & +
& 1\\ 2 & = & 2\times 1 & + & 0
\end{eqnarray}
To find the solution we use the last non-zero remainder,
namely $1=5-2[2]$
and successively substitute the remainders from the other
equations until we get back to the first one giving a combination
of the two original values $P=83$ and $Q=19$. The method in this
example has the following steps with the remainders given in
square brackets.
\begin{eqnarray} 1 & = & [5]-2[2]\\ & = &
[5]-2([7]-[5])\\ & = & -2[7]+3[5]\\ & = &
-2[7]+3(19-2[7])\\ & = & (3\times 19)-8[7]\\ & =
& (3\times 19)-8(83-(4\times 19))\\ & = & (-8\times
83)+(35\times 19). \end{eqnarray}
Thus a solution of $83x+19y=1$ is $x=-8$ and $y=35$.
Can you find a solution of $83x+19y=7$?
Can you now find a solution of $827x+191y=2$? You should first
solve the equation $827x+191y=1$ (using the computer if you
wish).
For the next article in the series, click here .