A history of Greek mathematics - Wilbourhall.org

INDETERMINATE ANALYSIS 501
described above (pp. 477-9). The problem is ' to divide
unity into two (or three) parts such that, if one and the
same given number be added to each part, the results are
all squares '.]
(V. 10. x + y — 1, x + a = u 2 , y + b = v 2 .
\V. 12. x+y+z— 1, x + a = 16 2 , y + b = v 2 , z + c = w 2 .
[These problems are like the preceding except that
different given numbers are added. The second of the
two problems is not worked out, but the first is worth
reproducing. We must take the particular figures used
by Diophantus, namely a = 2, b — 6. We have then to
divide 9 into two squares such that one of them lies
between 2 and 3. Take two squares lying between 2
2
and 3, say fJJ,
.
fff We have then to find a square £
lying between them ; if we can do this, we can make
9 — £
2
a square, and so solve the problem.
Put 9-£ 2 = (3-m£) 2 , say, so that £ = 6m/(m 2 + 1)
;
and m has to be determined so that £ lies between
T2 ancl
T§
•
rnu e 17 6m 19
Inereiore — < —^ < — •
12 m 2 +l 12
Diophantus, as we have seen, finds a fortiori integral
limits for m by solving these inequalities, making m not
greater than ff and not less than J§
(see pp. 463-5 above).
He then takes m = 3J and puts 9 — £
2
= (3 — 3J£)
2
which gives £ = f§.]
V. 13. x + y + z = a, y + z — u
2
, z + x = v 2 ,
x + y = w 2 .
,
V. 14. x + y + z + w = a, x + y + z = s 2 , y
+ z + w=t 2 ,
z + w + x — u
2i
w
+ x + y — v 2 .
[The method is the same.]
V. 21. x 2 y 2 z 2 + x 2 = u 2 , x 2 y 2 z 2 + y
2 = v
2, x 2 y 2 z 2 + z 2 = iv 2 .
V. 22. x 2 y 2 z 2 -x 2 = < x 2 y 2 z 2 -y 2 = v 2 , x 2 y 2 z 2 -z 2 = w 2 .
V. 23. x 2 ~x 2 y 2 z 2 = u 2 , y 2 -x 2 y 2 z 2 = v 2 , z 2 -x 2 y 2 z 2 = iv 2 .
[Solved by means of right-angled triangles in rational
numbers.]