\magnification=\magstep1
\def\Kappa{K}
\def\Mu{M}
\def\s{\varsigma}
\noindent
Epitaph of Diophantos of Alexandria (possibly 3rd century CE?):  
\bigskip
This tomb holds Diophantos.  Ah, how great a marvel!  The tomb tells scientifically the
measure of his life.  God granted to him to be a boy for the sixth part of his life,
and adding a twelfth part to this, his cheeks were clothed with down; he lit the 
light of wedlock after a seventh part, and five years after his marriage, he
was granted a son.  Alas! late-born wretched child; after attaining the measure of
half his father's life, chill Fate took him.  After consoling his grief with
the science of numbers for four years, he ended his life.  ({\it Greek Anthology}, 126)
\bigskip
\noindent
From: {\bf Book I} of the {\it Arithmetica} of Diophantos of Alexandria
\bigskip
\noindent
{\it (Not strictly literal) translation from the Greek by JL}
\bigskip
\noindent
Most honored Dionysius, knowing that you are eager to learn to [set up and]
solve problems concerning numbers, I tried starting from the foundations
that are the basis of these things to lay down the nature and the power
of numbers.

\bigskip
This thing might seem more difficult since it is not
yet familiar, for the souls of beginners do not hope to succeed, but
it will be easy for you because of your enthusiasm and my
exposition of the material; for
desire comes to knowledge quickly when it receives teaching.

\bigskip
But besides these things you know that all numbers are compounded
from some number of units, and their production has no limit.  There
happen to be among them:
\bigskip
\item{$\bullet$} {\it squares}, which are formed by multiplying
some number by itself, and this number is called the
side of the square;
\item{$\bullet$} {\it cubes}, which are formed from squares
by multiplying by their sides;
\item{$\bullet$} {\it power-powers}, which are formed by
multiplying a square by itself;
\item{$\bullet$} {\it power-cubes}, which are formed by from
squares by multipying by the cube with the same side;
\item{$\bullet$} {\it cubo-cubes}, which are formed by multiplying
a cube by itself.
\bigskip
It turns out that most arithmetical problems are woven by adding
or subtracting or multiplying or taking ratios of these with each
other or with the sides.  As you proceed along the path that will be shown
to you, they are solved.

Each of these numbers has been accepted to be an element of the theory
of arithmetic using an abbreviation.
\bigskip
\item{$\bullet$} The square numbers are called
powers and the sign of these is the $\Delta$ [capital ``delta'']
having an $\Upsilon$ [capital ``upsilon''] superscript:
$\Delta^\Upsilon$ -- ``power.''  [JL:  the Greek word is {\it dunamis}]
\item{$\bullet$} The symbol for a cube is $\Kappa$ with the $\Upsilon$
superscript: $\Kappa^\Upsilon$ -- ``cube.''  [JL: the Greek word is {\it kubos}]
\item{$\bullet$}  The symbol for a square multiplied by itself
is two capital deltas with the $\Upsilon$ superscript:  $\Delta^\Upsilon\Delta$ --
``power-power.''
\item{$\bullet$} The symbol for a square multiplied by the cube of its
side is $\Delta\Kappa$ with the $\Upsilon$ superscript: $\Delta\Kappa^\Upsilon$ --
``power-cube.''
\item{$\bullet$}  The symbol for a cube multiplied by itself is the
double $\Kappa$ with superscript $\Upsilon$: $\Kappa^\Upsilon\Kappa$ --
``cubo-cube.''
\item{$\bullet$}  The symbol for a number having none of these types, but having an
indeterminate number of units in itself is $\s$.
\item{$\bullet$}  There is another constant part of definite numbers, the unit,
and its symbol is $\Mu$ with superscript $o$: $M^o$ -- unit [JL: the Greek word is {\it monos}].
\bigskip
Just as the parts of numbers are named after the numbers\footnote{${}^1$}{JL: What is happening here is that Diophantos is giving names for the reciprocals of
these different classes of numbers.  For instance the reciprocal number 5 is the $1/5$th part, the reciprocal
of a square $x^2$ is a square part $1/x^2$ and so forth.}, so the parts of
the numbers just named will be given names derived from those
numbers as follows:
\bigskip
\item{} from the number---the numbered part
\item{} from the power---the powered part
\item{} from the cube---the cubed part
\item{} from the power-power---the power-powered part
\item{} from the power-cube---the power-cubed part
\item{} from the cubo-cube---the cubo-cubed part
\bigskip
Each of these will have, in addition to the corresponding symbol
of the number, the sign $\chi$ distinguishing its species [or type].

Now that I have set out for you the name of each of the numbers, I will
go on to give their multplications.  These will be almost clear from
the names:
\bigskip
\item{} A number multiplied by a number makes a power [i.e. a square]
\item{} A number multiplied by [its] power makes a cube
\item{} A number multiplied by [its] cube makes a power-power
\item{} A number multiplied by [its] power-power makes a power-cube
\item{} A number multiplied by [its] power-cube makes a cubo-cube
\item{} A power multiplied by a power makes a power-power
\item{} A power multiplied by a cube makes a power-cube
\item{} A power multipled by a power-power makes a cubo-cube
\item{} A cube multplied by a cube makes a cubo-cube.
\bigskip

A number multiplied multiplied by its part makes a unit.

Since the unit does not change and stands forever, any species
multiplied by it will have the same species.

When the parts having the same name are multiplied, they will
make parts having the same name as the numbers.  For instance,
a numbered part multiplied by:
\bigskip
\item{}  a numbered part makes a powered
part
\item{} [its] powered part makes a cubed part
\item{} [its] cubed part makes a power-powered part
\item{} [its] power-powered part makes a power-cubed part
\item{} [its] power-cubed part makes a cubo-cubed part
\bigskip
That is, it will come out as the names suggest.
\bigskip
A numbered part multiplied
\bigskip
\item{} by a power makes a number
\item{} by a cube makes a power
\item{} by a power-power makes a cube
\item{} by a power-cube makes a power-power
\item{} by a cubo-cube makes a power-cube
\bigskip
A powered part multiplied:
\bigskip
\item{} by a number makes a numbered part
\item{} by a cube makes a number
\item{} by a power-power makes a power
\item{} by a power-cube makes a cube
\item{} by a cubo-cube makes a power-power
\bigskip
A cubed part [multiplied]:
\bigskip
\item{} by a number makes a powered part
\item{} by a power makes a numbered part
\item{} by a power-power makes a number
\item{} by a power-cube makes a power
\item{} by a cubo-cube makes a cube
\bigskip
A power-powered part [multiplied]:
\bigskip
\item{} by a number makes a cubed part
\item{} by a power makes powered part
\item{} by a cube makes a numbered part
\item{} by a power-cube makes a number
\item{} by a cubo-cube makes a power
\bigskip
A power-cubed part [multiplied]
\bigskip
\item{} by a number makes a power-powered part
\item{} by a power makes a cubed part
\item{} by a cube makes a powered part
\item{} by a power-power makes a numbered part
\item{} by a cubo-cube makes a number
\bigskip
A cubo-cubed part [multiplied]
\bigskip
\item{} by a number makes a power-cubed part
\item{} by a power makes a power-powered part
\item{} by a cube makes a cubed part
\item{} by a power-power makes a powered part
\item{} by a power-cube makes a numbered part
\bigskip
A lack times a lack is a presence, a lack times a presence is a lack,
and the symbol for a lack is $\Lambda$ [JL: from the Greek word {\it leipsis}].

\bigskip
Now that you have been apprised of the multiplications, the divisions
of the species that have been set forth are clear.  As a beginner in this
business, it will
be well to practice with addition and subtraction and multiplication with
the species, and how you might add species, both present and lacking,
with other species (which may themselves be present or lacking).

\bigskip
After these things, if it happens in some problem that some species
are equal to the same species, but they are not of the same number,
it will be necessary to subtract similar things from each of the
two parts until one species becomes equal to one species.  If somehow
in either of the two parts or in both some lacking species are present,
it will be necessary to add the lacking species in both parts, until
the species of the two parts become present, and again to subtract
similar things from similar things, until one species is left behind
in each of the two parts.

\bigskip
Let this be practiced on what is set down in the propositions, if possible,
until one species is left equal to one species.  Later we will also
show you how problems where two species are left equal to one are solved.

\bigskip
Now let us be on the way to the propositions, since we have gathered
together much material on the species.  They are many in number and
great in bulk, and because of this they are established slowly for those
who study them.  Some of them are difficult to remember, so I decided
to split them up where possible, and above all to begin with those
that act as elements and proceed from the simpler to the more complex.
In this way they will be accessible to beginners and their training will
be more memorable, when their study in thirteen
books has been completed.

\bigskip
\noindent
{\bf Proposition 1.}  To divide the proposed number into two numbers
with a given difference.
\bigskip
Let the given number be $100$ and the difference be $40$.  To find the
numbers.  Let the lesser number be set out $\s 1$.  Then the larger
will be $\s 1 M^o 40$.  Both together will be $\s 2 M^o 40$.  Therefore
$M^o 100$ is equal to $\s 2 M^o 40$.  And from similar things take away
similar things.  I take away $M^o 40$ from the $M^o 100$ [and the $\s2 M^o 40$], and the remaining
$\s 2$ are equal to $M^o 60$.  Therefore each $\s$ is $M^o 30$.  To what
was set down:  The lesser number will be $M^o 30$ while the greater will be
$M^o 70$ and the demonstration is clear.

\bigskip
\noindent
{\bf Proposition 2.}  It is necessary to divide the proposed number
into to numbers in a given ratio.
\bigskip
Let it be proposed to divide the number $60$ into two numbers in
$3^{pl}$ [i.e. triple] ratio.  Let the lesser number be set out $\s 1$.
The the greater will be $\s 3$ and the greater is three times the lesser.
The remaining requirement is that the two numbers [together] be equal
to $M^o 60$.  But the two together are $\s 4$.  So $\s 4$ is equal
to $M^o 60$, and $\s$ equals $M^o 15$.  Then the smaller number is
$M^o 15$ and the larger is $M^o 45$.

\bigskip
\noindent
{\bf Proposition 3.}  To divide the proposed number into two numbers
in a given ratio and a given difference.
\bigskip
Let it be proposed to divide the number $80$ into two numbers such that
the greater number is $3^{pl}$ the lesser and still exceeds by $M^o 4$.
Let the lesser number be set out $\s 1$.  Then the greater is $\s 3$ and
$M^o4$.  The remaining thing I wish is that the two numbers together
are $M^o80$.  But the two together are $\s 4 M^o4$.  Therefore
$\s 4 M^o4$ and $M^o 80$ are equal.  And I take away similar things from
similar things.  It remains $M^o 76$ is equal to $\s 4$ and $\s$ will be
$M^o 19$.  To what was set down:  the lesser number will be $M^o 19$
and the greater will be $M^o 61$ [when the $M^o 4$ which I took away
from the $M^o 80$ are added.  For I took them away to find out how
many $M^o$ each number will be, and later I add the $M^o 4$ to the
larger number, after knowing how many are in each.]

\bigskip
\noindent
{\bf Proposition 4.}  To find two numbers in a given ratio so that
their difference is also given.
\bigskip
Let it be proposed that the larger number be $5^{pl}$ the smaller, and
to make their difference $M^o 20$.  Let the lesser number be set out
$\s 1$.  Then the larger will be $\s 5$.  The remaining thing I wish
is that $\s 5$ exceed $\s 1$ by $M^o 20$.  But their difference is
$\s 4$ and this equals $M^o 20$.  Therefore the smaller number is $M^o 5$
and the larger is $M^o 25$.  And the larger continues to be $5^{pl}$ the
smaller and the difference is $M^o 20$.

\bigskip
\noindent
{\bf Proposition 5.}  To divide the proposed number into two numbers
so that of each of the numbers obtained, the given non-identical parts
added together equal a given number.
\bigskip
It is clearly necessary for the given number to be between the two 
numbers that come about if the given non-identical parts of the proposed number are 
made.  
\bigskip
Let it be proposed to divide $100$ into two numbers so that of the first
number the 3rd part and of the second number the 5th part taken 
together are $M^o 30$.  Let the 5th part of the second number be $\s 1$;
the second number itself is then $\s 5$.  Then the 3rd part of the first
number is will be $M^o 30 \Lambda \s 1$, and the first number itself will
be $M^o 90 \Lambda \s 3$.  The remaining thing I wish is for the two
numbers together to equal $M^o 100$.  But the two numbers together are $M^o 90 \s 2$;
and these are equal to $M^o 100$.  And from similar things similar things. Therefore
$M^o 10$ will equal $\s 2$ and $\s$ will be $M^o 5$.
\bigskip
To what was set down.  I set out the 5th part of the second number $\s 1$.  This equals
$M^o 5$, and therefore the second number itself is $M^o 25$.  The 3rd part of the first 
is $M^o 30 \Lambda \s 1$, which is $M^o 25$, so the first number itself is $M^o 75$.  
And the 3rd part of the first number and the 5th part of the second number remain
$M^o 30$ [together they make the proposed number].    

\bigskip
\noindent
{\bf Proposition 6.}  To divide the proposed number into two numbers such that the given
part of the first exceeds the given part of the other by a given number.
\bigskip
It is clearly necessary that the given number is smaller than the number that comes about 
if the given part of the number proposed from the start in which the excess exists is taken.
\bigskip
Let it be proposed to divide $100$ into two numbers such that the 4th part of the first exceeds
the 6th part of the second by $M^o 20$.  
\bigskip
I set out the 6th part of the second, $\s 1$; the second number itself will be $\s 6$.  Then
the 4th part of the first will be $\s1 M^o 20$, and therefore the first number itself will be
$\s 4 M^o 80$.  The remaining thing I wish is that the two numbers added together give $M^o 100$.
But the two numbers added together make $\s 10 M^o 80$ and these are equal to $M^o 100$.  From
similar things similar things.  What remains is $\s 10$ is equal to $M^o 20$ and the $\s 1$ becomes
$2$.
\bigskip
To what was set down.  I set out the the 6th part of the second number $\s 1$; it will be $M^o 2$
and therefore the number itself will be $M^o 12$.  The 4th part of the first is $\s 1 M^o 20$, so 
it is $M^o 22$, and therefore the first number itself is $M^o 88$.  And the 4th part of the first
number continues to exceed the 6th part of the second by $M^o 20$ and the two numbers together 
make the proposed number.  

\bigskip
\noindent
From:  {\bf Book II} of the {\it Arithmetica}
\bigskip
\noindent
{\bf Proposition 8.} To divide the proposed square into two square numbers.
\bigskip
Let it be proposed to divide the number $16$ into two square numbers.  Let the first
number be set out $\Delta^\Upsilon 1$.  Therefore the other number will be $M^o 16 \Lambda \Delta^\Upsilon 1$.
It will therefore be necessary for $M^o 16 \Lambda \Delta^\Upsilon$ to be equal to a square.
I form the square from however many $\s$'s, $\Lambda$ as many $M^o$ as is the side of 
$M^o 16$, let it be $\s 2 \Lambda M^o 4$.  Then the square itself will be $\Delta^\Upsilon 4 M^o 16 \Lambda \s 16$.
These are equal to $M^o 16 \Lambda \Delta^\Upsilon 1$.  Let the lack be added to both, and from similar things
similar things.

\bigskip
Therefore $\Delta^\Upsilon 5$ are equal to $\s 16$ and the $\s$ becomes the 5th part of $16 M^o$. 

\bigskip
The one number will be $256^{\backslash 25}$ [JL:  This what our existing manuscripts write for
the 25th part of 256, or ${256\over 25}$] and the other number will be $144^{\backslash 25}$ [JL: That is, 
${144\over 25}$.] ...
\bigskip
\noindent
Comments:
\bigskip
\item{1.}  This is one of the few times when Diophantos uses rational numbers instead of 
integers in the solution of one of his problems.
\item{2.}  John of Chortasmenos (1370 - 1437 CE) wrote the following marginal remark to this 
proposition ``Thy soul, Diophantos, be with Satan because of the difficulty of your theorems.''
\item{3.}  In 1621 CE, Pierre de Fermat wrote in his margin at this point: ``If an integer
$n$ is greater than $2$, then $a^n + b^n = c^n$ has no solutions in nonzero integers $a,b,$ and $c$.
I have a truly marvelous proof of this proposition which the margin is too narrow to contain.''  This
celebrated claim, which came to be known as {\it Fermat's Last Theorem}, was finally proved in the 
mid-1990's by Andrew Wiles and Richard Taylor using the theory of elliptic curves(!)
\end




