History of Hindu Mathematics: A Source Book v2.c17.s14.53
History of Hindu Mathematics: A Source Book v2.c17.s14.53
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v2.c17.s14.53
We have $Nα²+k=β²,$ $Nα'²+k'=β'².$ Multiplying the first equation by β'², we get $Nα²β'²+kβ'²=β²β'².$
History of Hindu Mathematics: A Source Book v2.c17.s14.53
We have $Nα²+k=β²,$ $Nα'²+k'=β'².$ Multiplying the first equation by β'², we get $Nα²β'²+kβ'²=β²β'².$