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Question Number 199112 by hardmath last updated on 28/Oct/23

$$\{\begin{array}{l}{\mathrm{a}}^2-\mathrm{b}=73\\ {\mathrm{b}}^2-\mathrm{a}=73\end{array}\mathrm{find}:\mathrm{a},\mathrm{b}=?$$

Answered by mr W last updated on 28/Oct/23

$$\left(i\right)-\left(ii\right):$$$$a^2-b^2+a-b=0$$$$(a+b+1)(a-b)=0$$$$case1:a-b=0$$$$a=b$$$$b^2-b-73=0$$$$\Rightarrow b=\frac{1\pm \sqrt{293}}{2}=a$$$$case2:a+b+1=0$$$$a=-b-1$$$$b^2+b-72=0$$$$b=\frac{-1\pm 17}{2}=8or-9$$$$a=-9or8$$

Commented by hardmath last updated on 28/Oct/23

$$\mathrm{thankyou}\mathrm{professor}$$

Answered by Frix last updated on 28/Oct/23

$$x^2-y=z$$$$y^2-x=z$$$$\Rightarrow$$$$y=z-x^2$$$$x^4-2{zx}^2-x+z^2-z=0$$$$(x^2-x-z)(x^2+x-z+1)=0$$$$\Rightarrow$$$$x=y=\frac{1\pm \sqrt{4z+1}}{2}$$$$x=\frac{-1\pm \sqrt{4z-3}}{2}\wedge y=\frac{-1\mp \sqrt{4z-3}}{2}$$

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