History of Hindu Mathematics: A Source Book v2.c17.s15.19

v2.c17.s15.19

&= a_1 (N + kk_1) - k_1 an, \\ & \ [\because\ \ kk_1 = n² - N] \\ &= a_1 N + k_1 b, \ [\because\ \ a_1 k = an + b] \\ \therefore b &= (b_1 n - N a_1)/k_1. \end{aligned}