Page:Indian mathematics, Kaye (1915).djvu/28

 (14) $$ax+1=s^2,\quad bx+a=t^2$$ (15) $$2(x^2-y^2)+3=s^2,\quad 3(x^2-y^2)+3=t^2$$ (16) $$ax^2+by^2=s^2,\quad ax^2-by^2+1=t^2$$ (17) $$x^2+y^2\pm 1=s^2,\quad x^2-y^2\pm 1=t^2$$ (18) $$x^2-a\equiv x^2-b\equiv o~Mod.~c$$ (19) $$ax^2+b\equiv o~Mod.~c$$ (20) $$x+y=s^2,\quad x-y=t^2,\quad xy=u^3$$ (21) $$x^3+y^3=s^2,\quad x^2+y^2=t^3$$ (22) $$x-y=s^3,\quad x^2+y^2=t^3$$ (23) $$x+y=s^2,\quad x^3+y^2=t^2$$ (24) $$\scriptstyle{x^3+y^2+xy=s^2,\quad (x+y)s+1=t^2}$$ (25) $$\scriptstyle{ax+1=s^3,\quad as^2+1=t^2}$$ (26) $$\scriptstyle{wxyz=a(w+x+y+z)}$$ (27) $$\scriptstyle{x^3-a\equiv o~Mod.~b}$$ (28) $$\begin{align}\\&\scriptstyle{x+y+3=s^2,~x-y+3=t^2,~x^2+y^2-4=u^2,}\\&\scriptstyle{x^2-y^2+12=v^2,\ \frac{xy}{2}+y=w^3,}\\&\scriptstyle{s+t+u+v+w+2=z^2}\\\end{align}$$ (29) $$\scriptstyle{\frac{\sqrt[3]{xy+y}}{2}+\sqrt{x^2+y^2}+\sqrt{x+y+2}+\sqrt{x-y+2}+\sqrt{x^2-y^2+8}=t^2}$$ (30) $$\begin{align}\\&\scriptstyle{w+2=a^2,~x+2=b^2,~y+2=c^2,~z+2=d^2,}\\&\scriptstyle{wx+18=e^2,\ xy+18=f^2,\ yz+18=g^2,}\\&\scriptstyle{a+b+c+d+e+f+g+11=13}\\\end{align}$$. 14. —The Indian mathematicians of this period seem to have been particularly attracted by the problem of the rational right-angled triangle and give a number of rules for obtaining integral solutions. The following summary of the various rules relating to this problem shows the position of the Indians fairly well.