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Question Number 141246 by ajfour last updated on 17/May/21

$$r=q+1$$$$pq=q+1$$$$c^{2}p={qr}^2,helpfindp,q,r.$$

Answered by Rasheed.Sindhi last updated on 17/May/21

$$r=pq$$$$c^{2}p={qr}^2\Rightarrow c^{2}p=q{\left(pq\right)}^2$$$$\Rightarrow c^{2}p-p^2{q}^3=0$$$$p(c^2-{pq}^3)=0$$$$p=0\mid c^2={pq}^3$$$${}^{\bullet }p=0\Rightarrow q=-1\Rightarrow r=0$$$$(p,q,r)=(0,-1,0)$$$${}^{\bullet }{c}^2={pq}^3\Rightarrow p=\frac{c^2}{q^3}$$$$pq=q+1\Rightarrow \left(\frac{c^2}{q^3}\right)q=q+1$$$$q^3+q^2-c^2=0$$$$.....$$

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