VS. 18] ANOTHER ASTRONOMICAL PROBLEM 101 where s=0, 1,2 i=0 gives the least solution. 1 Another astronomical problem on indeterminate equations : 18. The residue of the minute of Mars multiplied by the cube of two and increased by one yields a square-root (without remainder) ; that square number multiplied by seven and then further increased by one is again a perfect square. Having as- certained the residue from this (hypothesis) one who can find out ihe longitude of Mars and the ahargana together with the number of solar years elapsed is (indeed) the foremost amongst the intelligent mathematicians on this earth girdled by the oceans. Let x denote the residue of the minute of Mars, trten we have to solve the equations 8*+ !=/, say, (i) 7y4-l=* 2 , say, (2) Eliminating j> between (1) and (2), we get, 56*+8=c 2 . (3) Evidently *=1, £=8 is a solution of this equation, so that we may take 1 as the residue of the minute for Mars. 2 Let « be the ahargana corresponding to this residue of Mars. Then 165371328m- 1 Ml =o, (4) 5259725 v ; where c denotes the revolutions performed by Mars is u days. Solving (4) we get « = 1863192 days, © = 2712 revolutions, sign, 25°, 31', which agrees with the solution given by the commentator Sankaranarayana. The commentator Sankaranarayana has also given an alternative inter- pretation of the text. According to that interpretation the above stanza 1 The results obtained above agree with those given by the commentator Sankaranarayana. It may be noted that there is no ahargana which may satisfy both equations (1) and (2) above. For, if we take u=£, then we get 39447937f«-131493125.r+228612794*0, which is impossible. 2 According to the commentator Udaya Divakara, one should first find the value of y by solving (2) and then substituting this value in (1) find x.
