A history of Greek mathematics - Wilbourhall.org

514 DIOPHANTUS OF ALEXANDRIA
term in the first). Diophantus takes p = 64, making
the equations
m 2 — 61 44 m + 1048576 = a square]
m + 64 = a square)
Multiplying the second by 16384, and subtracting the two
expressions, we have as the difference m 2 — 22528m.
Diophantus observes that, if we take m, m— 22528 as
the factors, we obtain m = 7680, an impossible value for
the area of a right-angled triangle of perimeter
VI. 24. z = u +
u, x = v 3 — v, y = w*.
[VI. 6, 7]. (ixf+%mxy = u 2 .
[VI. 8, 9]. H(x + y)}
2
+ ±mxy = u 2 .
[VI. 10, 11]. {i(z + x)} 2 + ±mxy = u 2 .
[VI. 12.] y + (x—y).%xy = u 2 , x = v 2 . (x > y.)
[VI. 14, 15], u 2 zx — \xy .x(z—x) = v 2 . (u 2 < or > \xy.)
The treatise
on Polygonal Numbers.
The subject of Polygonal Numbers on which Diophantus
also wrote is, as we have seen, an old one, going back to the