The History of Mathematical Proof in Ancient Traditions

320 ian mueller

multiplied by a certain number which, when a unit is added to it, is double of the

multitude of all the numbers set out with the unit.

if p = x 1 + x 2 +... + xn, where xi + 1 = xi + j and x 1 = 1, then

p 8 j + (j − 2)^2 = ((2n − 1)j + 2)^2 [= ((n + n − 1)j + 2)^2 ].

It is easy to prove Dioph 4 using Dioph 2 and 3, 14 as Diophantus does,

although his argument is cumbersome. Here I wish only to present the

very beginning of his argument and a diagram, provided by me, represent-

ing it.

[460,13] For let AB, CD, EF be numbers in equal excess starting from the unit. 15

I say that the proposition results. For let there be as many units in GH as the

numbers set out with the unit. And since the excess by which EF exceeds a unit

is the excess by which AB exceeds a unit multiplied by GH minus 1, 16 if we set

out each unit, AK, EL, GM, we will have that LF is KB multiplied by MH. So LF

is equal to the product of KB, MH. And if we set out KN as 2, we will investigate

whether, if the sum is multiplied by 8 KB (which is their excess) and the square of

NB (which is less than their excess by 2) is added, the result is a square of which the

side minus 2 produces a number which is their excess (KB) multiplied by GH,HM

together ( Figure 9.5 ).

Figure 9.5 Diophantus’ diagram, Polygonal Numbers, Proposition 4.

AK N B x 2

12 j −^2

KB j

EL F xn

1 xn − 1

GM H n

1 n − 1

14 Since:

((2n −1)j + 2)^2 − ( j −2)^2

= 4 n^2 j^2 − 4 nj^2 + j^2 + 8 nj − 4 j + 4 − j^2 + 4 j − 4

= 4 n^2 j^2 − 4 j^2 n + 8 nj = 4 j(n^2 j − jn + 2 n),

to prove Dioph 4 we need only prove:

2(x 1 + x 2 +... + xn) = n^2 j − jn + 2 n,

or, by Dioph 3:

(xn + x 1 )n = n^2 j − jn + 2 n, that is xn+ x 1 = nj − j + 2.

15 Note that AB, CD, and EF are numbers, not the unit.

16 Cf. Dioph 2.