A history of Greek mathematics - Wilbourhall.org

I III.
494 DIOPHANTUS OF ALEXANDRIA
}
Then, since the sum of the first three expressions is
itself equal to x + y + z, we have
x = i(6f + c 2 ), y = 2
i(c + a 2 ), z = i(«2 + & 2 ).]
x + y + a — w 2 .
, z + x — a = v 2 ,
[Suppose yz + a = m 2 , and let y = (m 2 — a)£, z = 1
/£:
also let zx + a = ti 2 ;
therefore a; = (w — 2 a)£.
We have therefore to make
(m — 2 a) (n
2 — 2 a) g + a a square.
Diophantus takes m 2 =25, a = 12, w, = 2 16, and
arrives at 52£ 2 +12, which is to be made a square.
Although 52 . 1 2 + 12 is a square, and it follows that any
number of other solutions giving a square are possible
by substituting 1+rj for g in the expression, and so on,
Diophantus says that the equation could easily be solved
if 52 was a square, and proceeds to solve the problem of
finding two squares such that each increased by 12 will
give a square, in which case their product also will be
a square. In other words, we have to find m 2 and n 2
such that m 2 — a, n 2 — a are both squares, which, as he
says, is easy. We have to find two pairs of squares
differing by a. If
a = pq=p'q', {% (p-q)}* + a = {i(p + q)}\
and {h'(p'-q')} 2 2
+a = {Hp+q')} ;
let, then, m 2 2
= { \ (p + q)
III. 6. x + y + z = t 2 , y + z = u 2 , z + x = v 2 ,
x + y — w 2 .
III. 7. x — y — y — z, y + z — u 2 ,
z + x = v 2 , x + y = w 2 .
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III. 8. x + 2/ + z +