A history of Greek mathematics Vol.II from Aristarchus to Diophantus by Heath, Thomas Little, Sir, 1921

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
2/ ) + - 2 (y ) + (x 2 ~z 2 ). (x>y> z)
IV. 16. x + 2/ 4- z — t 2 ,
x 2 + 2/
= '^ 2
>
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
numbers are therefore 13?? 2 — 1, 52?7 2 , 104?7 2 -f-l, and
,
; the
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
498 DIOPHANTUS OF ALEXANDRIA
Similarly we satisfy the second condition by assuming
zx — n
2
£
2
+ 2ng ; therefore x = n 2 g + 2 n. To satisfy the
third condition, we must have
(m 2 n 2 £ 2 + 2mn .
m
+ n£ + 4m%) + 1 a square.
We must therefore have 4 mn + 1 a square and also
mn(m + n) = mn V(4mn-\- 1). The first condition is
satisfied by n = m-\- 1 , which incidentally satisfies the
second condition also. We put therefore yz = (m£ + l) 2 — 1
2
and zx— {
(m + l)£ + 1 }
— 1, and assume that z = g, so that
y = m 2 g + 2m, x = (m + 1) 2 £ + 2(m + 1), and we have
shown that the third condition is also satisfied. Thus we
have a solution in terms of the undetermined unknown £.
The above is only slightly generalized from Diophantus.]
IV. 20. x 2
x z
-\- 1 = r 2 , x
z
x Y
-\~ 1 = s 2 ,
x
X-i X/i ~f~ 1 — iXi , Xc) Xi ~t~ — J-
1
x 2
+ 1 = t 2 ,
U ,
Xr, Xa ~t" i — w .
[This proposition depends on the last, x lt
x 2
, x z
being
determined as in that proposition. If x z
corresponds to z
in that proposition, we satisfy the condition x 3
x 4
+l = w 2
by putting x 3
x± = {(m + 2)£ + 1 }
2 — 1, and so find x 4 in
terms of £, after which we have only two conditions more
to satisfy. The condition x 1
x + 1 = square is automatically
satisfied,
4:
since
f
(m + 1)
2
£ + 2 (m + 1)} {
(m + 2)
2
£ + 2 (m + 2) } + 1
is a square, and it only remains to satisfy x 2
x i
+l = square.
That is,
(m 2 |+2m) {(m + 2)
2
£ + 2(m + 2)} + 1
= m 2 (m+2) 2 f + 2m(m + 2)(2m + 2)^ + 4m(m+2) + 1
has to be made a square, which is easy, since the coefficient
2
of £ is a square.
With Diophantus m = 1, so that x x
= 4£ + 4, x 2
= £ + 2,
# 3
= £, # 4
= 9£ + 6, and 9£ 2 + 24£+13 has to be made
a square. He equates this to (3£— 4) 2 ,
giving £ = y
1^.]
IV. 21. xz = y
2
, x — y = u 2 , x — z = v 2 , y
IV. 22. xyz + x = u 2 , xyz + y — v 2 , xyz + z = w 2 .
IV. 23. xyz — x = u 2 , xyz — y = i; 2 ,
xyz — z = iv 2 .
— z — w 2 . (x>y>z)