ब्राह्मस्फुटसिद्धान्त (ब्रह्मगुप्त - शून्य, कुट्टक, बीजगणित एवं सम्पूर्ण २१ अध्याय सान्वय सटीक)

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SOLUTION IN POSITIVE INTEGERS 255 (ii) Let k=±2; then the auxiliary equation is Nα²±2=β² By Brahmagupta's Corollary, we have N(2αβ)²+4=(β²+Nα²)² or N (αβ)²+1=( (β²+Nα²)/2 )² Hence the required first solution is x=αβ, y=½(β²+Nα²) Since Na²=β² ∓ 2, we have ½ (β²+Nα²)=β² ∓ 1=a whole number. (iii) Now suppose k=+4: so that Nα²+4=β² With an auxiliary equation like this, the first integral solution of the equation Nx²+1=y² is x=½αβ y=½(β²-2); if α is even; or x=½