ब्राह्मस्फुटसिद्धान्त (ब्रह्मगुप्त - शून्य, कुट्टक, बीजगणित एवं सम्पूर्ण २१ अध्याय सान्वय सटीक)

PRELIMINARY OPERATIONS 229
putting similarly x=q₁y₁+x₁
the equation (iii) can be further reduced to
r₁x₁=r₂y₁—c (iv)
and so on.
Writing down the successive values and reduced equations
in columns, we have
(1) y=qx+y₁ (I.1) by₁=r₁x+c
(2) x=q₁y₁+x₁ (I.2) r₁x₁=r₂y₁—c
(3) y₁=q₂x₁+y₂ (I.3) r₂y₂=r₃x₁+c
(4) x₁=q₃y₂+x₂ (I.4) r₃x₂=r₄y₂—c
(5) y₂=q₄x₂+y₃ (I.5) r₄y₃=r₅x₂+c
(6) x₂=q₅y₃+x₃ (I.6) r₅x₃=r₆y₃—c
......... .........
(2n-1) yₙ₋₁=q₂ₙ₋₂ xₙ₋₁+yₙ (I. 2n-1) r₂ₙ₋₂ yₙ=r₂ₙ₋₁ xₙ₋₁+c
(2n) xₙ₋₁=q₂ₙ₋₁ yₙ+xₙ (I. 2n) r₂ₙ₋₁ xₙ=r₂ₙ yₙ—c
(2n+1) yₙ=q₂ₙ xₙ+yₙ₊₁ (I. 2n+1) r₂ₙ yₙ₊₁=r₂ₙ₊₁ xₙ+c
Now the mutual division can be continued either (i) to the
finish or (ii) so as to get a certain number of quotients and
then stopped. In either csse the number of quotients found, negle-
cting the first one (q), as is usual with Āryabhaṭa, may be even
or odd.
Case (i) First suppose that the mutual division is continued
until the zero remainder is obtained. Since a, b are prime to
each other, the last one remainder is unity.
Subcase (i.1.). Let the number of quotients be even. We
then have
r₂ₙ=1, r₂ₙ₋₁=0, q₂ₙ=r₂ₙ₋₁
The equations (1,2n) and (I.2n+1), therefore become
yₙ=q₂ₙ xₙ+c
and
yₙ₊₁=c
respectively. Giving an arbitrary integral value (t) to xₙ
we get an integral value of yₙ. From that we can find the value
of xₙ₋₁ by the equation (2n). Procceding backwards step by
step we ultimately find the values of x and y in positive integers.
So that the equation (I) is solved.
Subcase (i. 2) : If the number of quotients be odd, we shall
have
r₂ₙ₋₁=1, r₂ₙ=0, q₂ₙ₋₁=r₂ₙ₋₂.