History of Hindu Mathematics: A Source Book v1.c8.s10.12

v1.c8.s10.12

Multiplying the congruences, we obtain $n.n₂ n_p n′₁.n′₂ n′_p 9.$ Further let $R r′ 9$ Therefore $n₁.n₂.n₃ n_p± R n′₁.n′₂ n′_p ± r′ 9.$ Hence $N n′₁.n′₂ n′_p ± r′ 9.$ In particular, if $n₁ = n₂ = = n_p = n, say$ then will $n′₁ = n′₂ = = n′_p = n′.$ Therefore, $N = np ± R$ and $N n′ p ± r′ 9.$