History of Hindu Mathematics: A Source Book v2.c17.s14.143

Now, by the Lemma, we obtain from (2) and (3) N{1/2\alpha\beta(\beta² &+ 3)(\beta² + 1)}² + 1 / &= [(\beta² + 2){1/2(\beta² + 3)(\beta² + 1) - 1}]².