DETERMINATE EQUATIONS 
489 
whence x + 1 = 
'si 
U 
(/3+1) (y +1) 
a + 1 
and it is only necessary that (a + 1) (/3+1) (y +1) should 
be a square, not that each of the expressions a + 1, (3 + 1, 
y + 1 should be a square. 
Dioph. finds in a Lemma (see under (vi) below) a solu 
tion kv dopiarco (indeterminately) of* xy+(x + y) = k, 
which practically means finding y in terms of ic.] 
IV. 35. yz—(y + z) — a 2 — 1, zx — (z + x) = h 2 — 1, 
xy—(x + y) = c 2 — 1. 
[The remarks on the last proposition apply mutatis 
mutandis. The lemma in this case is the indeterminate 
solution of xy — (x + y) = k.] 
IV. 37. yz — a(x + y + z), zx = b(x+y + z), xy — c(x + y + z). 
[Another interesting case of ‘ false hypothesis ’. Dioph. 
first gives x + y + z an arbitrary value, then finds that 
the result is not rational, and proceeds to solve the new 
problem of finding a value of x + y + z to take the place of 
the first value. 
If w = x +y +z, we have x — cw/y, z — aw/y, so that 
zx = acvj 2 / y l — bw by hypothesis ; therefore y 2 = w. 
For a rational solution this last expression must be 
a square. Suppose, therefore, that w — ~~ ¿ 2 , and we have 
x + y + z = -j£ 2 , y = -^i, z = a£, x = c£. 
Eliminating x, y, z, we obtain £ = (bc + ca + ab)/ac, 
and 
x = [be + ca + ab)/a, y = (be+ ca + ah)/ b, 
z 
= (be + ca + ab) / c.] 
Lemma to V. 8. yz = a 2 , zx = b 2 , xy = c 2 .