514 
UIOPHANTUS OF ALEXANDRIA 
term in the first). Diophantus takes p = 64, making 
the equations 
m 2 —6144m + 1048576 = a square 
m + 64 = a square 
Multiplying’ the second by 16384, and subtracting the two 
expressions, we have as the difference m 2 — 225 28m. 
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 p = 64. 
We therefore take as factors 11m, yym — 2048, and, 
equating the square of half the difference ( = fym+ 1024) 
to 16384m + 1048576, we have m = -f |t~. 
(3) Returning to the original problem, we have to 
substitute this value for m in 
and we obtain 
78848 £ 2 — 8432£ + 225 = 0, 
the solution of which is rational, namely £ = ■£-£-§ (or T f^). 
Diophantus naturally takes the first value, though the 
second gives the same triangle.] 
VI. 23. z 2 = xi 2 + u, z 2 /x — v z + v. 
VI. 24. z — u 3 + u, x = v 3 — v, y = iv 3 . 
[VI. 6, 7]. {^x) 2 + ^vixy = u 2 % 
[VI. 8, 9]. (% + y)} 2 + \ r mxy = u 2 . 
[VI. 10, 11]. {%{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