History of Hindu Mathematics: A Source Book v2.c17.s16.70

v2.c17.s16.70

(4) From these fractional roots, we may derive *integral roots* by the Cyclic Method. Since $13(½)² - 1 = (3/2)²,$ we have, by Bhâskara's Lemma, m being an indeterminate multiplier, $13((m/2 + 3/2)/(-1))² + (m² - 13)/(-1) = ((3m/2 + 13/2)/(-1))²,$ or $13((m + 3)/(-2))² + (m² - 13)/(-1) = ((3m + 13)/(-2))².$ The suitable value of m which will make (m+3)/2 an integer and |m² - 13| minimum is 3. So that we have $13.3² + 4 = 11².$ From this again we get the relation $13((3n + 11)/4)² + (n² - 13)/4 = ((11n + 13.3)/4)².$