History of Hindu Mathematics: A Source Book v2.c17.s14.131
History of Hindu Mathematics: A Source Book v2.c17.s14.131
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v2.c17.s14.131
$Since Nα²+4=β², (1)$ $we have N((α)/2)²+1=((β)/2)². (2)$
History of Hindu Mathematics: A Source Book v2.c17.s14.131
$Since Nα²+4=β², (1)$ $we have N((α)/2)²+1=((β)/2)². (2)$