History of Hindu Mathematics: A Source Book v2.c17.s15.5

v2.c17.s15.5

The *rationale* of this Lemma is simple: We have $Na²+k=b²,$ $and N.1²+(m²-N)=m², identically.$ Then by Brahmagupta's Lemma, we get $N(am+b)²+k(m²-N)=(bm+Na)².$ $∴ N((am+b)/k)²+(m²-N)/k=((bm+Na)/k)².$