Mathematics in Ancient and Medieval India c7.s5.166

c7.s5.166

and N. 1² + (m² - N) = m² then by *bhāvanā*, Bhāskara II obtained $N((am+b)/k)²+(m²-N)/k=((bm+Na)/k)²......(1)$ Next to choose the value of the m so that m²-N is numerically smaller and (am+b)/k becomes integer, since its value is determined by means of *kuṭṭaka*. $Let (am+b)/k=a₁, (m²-N)/k=k₁$ $(bm+Na)/k=b₁$