Polygonal numbers in ancient Greek mathematics 319
of their content; in footnotes I give simple algebraic proofs of the results
in order to show that the results are correct. Diophantus’ arguments are
cumbersome and roundabout.
[452,2] [Dioph 1] If three numbers exceed one another by an equal amount, then
eight times the product of the greatest and the middle one plus the square on the
least produces a square the side of which is equal to the sum of the greatest and
twice the middle one.
if x = y + k and y = z + k, then 8xy + z^2 = (x + 2 y)^2.^11
[454,6] [Dioph 2] If there are numbers in any multitude in equal excess, <the
excess> of the greatest over the least is their excess multiplied by one less than the
multitude of numbers set out.
if x 1 , x 2 ,... xn+ 1 are such that xi+ 1 = xi + j then xn+ 1 − x 1 = nj.^12
[456,2] [Dioph 3] If there are numbers in any multitude in equal excess, the sum
of the greatest and least multiplied by their multitude makes a number which is
double of the sum of the numbers set out.
if x 1 , x 2 ,... xn are such that xi+ 1 = xi + j then
(xn + x 1 )n = 2(x 1 + x 2 +... + xn).^13
[460,5] [Dioph 4] If there are numbers in any multitude in equal excess starting
from the unit, then the sum multiplied by eight times their excess plus the square of
two less than their excess is a square of which the side minus 2 will be their excess
I do not mean to suggest, nor do I believe, that Diophantus’ reasoning does not include
geometric elements of the kind we fi nd in the so-called geometric algebra of Book 2 of Euclid’s
Elements. But discussion of that issue would require a detailed examination of Diophantus’
proofs, a task which I cannot undertake here.
11 Proof: Let x = z + 2 k and y = z + k. Th en we should prove that:
8(z + 2 k)(z + k) + z^2 = ((z + 2 k) + 2(z + k))^2.
But
8 ( z + 2 k ) ( z + k ) + z 2 = 8 ( z^2 + 3 zk + 2 k^2 ) + z^2 = 9 z^2 + 24 zk + 16 k^2 =
( 3 z + 4 k)^2 = ( ( z + 2 k ) + 2 z + 2 k )^2 = ( ( z + 2 k ) + 2 (z + k ) )^2.
12 Dioph 2 is suffi ciently obvious that there is really nothing to prove, the basic idea being that
x 2 = x 1 + j, x 3 = x 2 + j = x 1 + 2 j, x 4 = x 3 + j = x 1 + 3 j , and so on.
13 We give an inductive proof of Dioph 3. For n = 1 the theorem says that x 2 − x 1 = 1. j. Suppose
(x 1 + xn)n = 2(x 1 + x 2 +... + xn). We wish to show that:
(x 1 + xn+ 1 )(n + 1) = 2(x 1 + x 2 +... + xn+ 1 ).
But:
(x 1 + xn+ 1 )(n + 1) = (x 1 + xn + j)(n + 1)
= (x 1 + xn)n + x 1 + xn + (n + 1)j = 2(x 1 + x 2 +...^ + xn) + x 1 + nj + xn + j
= 2(x 1 + x 2 +... + xn) + xn+ 1 + xn+ 1 = 2(x 1 + x 2 +... + xn+ 1 ).
(Th at xn+ 1 = x 1 + nj is a trivial reformulation of Dioph 2.)