A history of Greek mathematics - Wilbourhall.org

INDETERMINATE ANALYSIS 497
Put therefore x x
+ x 2
+ x 3
+ x± = 65 £.
and x 1
= 2. 39. 52 £
2
, ;r 2
= 2 . 25.60£
x i
= 2.16.63£ 2 ;
2 ,
cc 3
= 2. 33. 56 £
2
this gives 12768f = 65£, and £ = T^¥ -]
(IV. 4. x 2 + y = u 2 , x + y = u.
tlV. 5. £c 2 + 2/
= Uj x + y = u 2 .
IV. 13. $ + 1 = t 2 , y+1 = u 2 , x + y + 1 = v — 2
, 2/ a; + 1 = to 2 ,
[Put a; = (mg + l) 2 — 1 = m 2 £ 2 + 2m^; the second and
third conditions require us to find two squares with x as
difference. The difference m 2 £ 2 + 2m g is separated into
the factors m 2 £ + 2m, £; the square of half the difference
= {J(m,
2 — l)£ + m} 2 . Put
this equal to y+1, so
that y = i(m 2 — l)
2
£
2
+ m(m 2 — 1) £ + m 2 — 1, and the
first three conditions are satisfied. The fourth gives
J (m 4 — 6m 2 4- 1 ) £ 2 + (m 3 — 3 m) | + m 2 = a square, which
we can equate to (n£ — m) 2 .]
IV. 14. ^2 +
2
2/
+ z = 2 2
(^ - 2
2/
) + (y
2 - z
IV. 16. x + 2/ 4- z — t 2 ,
x 2 + 2/
= '^ 2
2
) + (x 2 ~z 2 ). (x>y> z)
>
y 2 + z = v 2 , z 2 + x = %v 2 .
[Put 4m£ for 7/, and by means of the factors 2m£, 2
we can satisfy the second condition by making x equal
to half the difference, or m£ — 1. The third condition
is satisfied by subtracting (4m£) 2 from some square, say
(4m£+l) 2 ; therefore z = 8mg+l. By the first con-
s
dition 13m£ must be a square. Let it be 169 77 ; the
numbers are therefore 13?? 2 — 1, 52?7 2 , 104?7 2 -f-l, and
the last condition gives 10816 ?;
4
+ 221 ??
2
= a square,
i.e. 10816t7 2 + 221 = a square = (104?? + l) 2 , say. This
gives the value of 77,
and solves the problem.]
IV. 17. x + y + z — t 2 , x 2 — y — u
2
,
y 2 — z — v 2 , z 2 — x = w 2 .
,
IV. 19. yz+1 = u 2 , zx + 1 = v 2 , xy+l
= iv 2 .
[We are asked to solve this indeterminately {kv tco
dopi). Put for yz some square minus 1, say m 2 £ 2
+ 2m£; one condition is now satisfied. Put z = £, so
thatj^/ = wi 2 £ + 2 m.
1523.2 K k