Question Number 141246 by ajfour last updated on 17/May/21
$${r}={q}+\mathrm{1} \\ $$$${pq}={q}+\mathrm{1} \\ $$$${c}^{\mathrm{2}} {p}={qr}^{\mathrm{2}} \:\:\:,\:{help}\:{find}\:{p},\:{q},\:{r}. \\ $$
Answered by Rasheed.Sindhi last updated on 17/May/21
$${r}={pq} \\ $$$${c}^{\mathrm{2}} {p}={qr}^{\mathrm{2}} \Rightarrow{c}^{\mathrm{2}} {p}={q}\left({pq}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{c}^{\mathrm{2}} {p}−{p}^{\mathrm{2}} {q}^{\mathrm{3}} =\mathrm{0} \\ $$$$\:\:\:\:\:{p}\left({c}^{\mathrm{2}} −{pq}^{\mathrm{3}} \right)=\mathrm{0} \\ $$$${p}=\mathrm{0}\:\mid\:{c}^{\mathrm{2}} ={pq}^{\mathrm{3}} \\ $$$$\:^{\bullet} {p}=\mathrm{0}\Rightarrow{q}=−\mathrm{1}\Rightarrow{r}=\mathrm{0} \\ $$$$\left({p},{q},{r}\right)=\left(\mathrm{0},−\mathrm{1},\mathrm{0}\right) \\ $$$$\:^{\bullet} {c}^{\mathrm{2}} ={pq}^{\mathrm{3}} \Rightarrow{p}=\frac{{c}^{\mathrm{2}} }{{q}^{\mathrm{3}} } \\ $$$${pq}={q}+\mathrm{1}\Rightarrow\left(\frac{{c}^{\mathrm{2}} }{{q}^{\mathrm{3}} }\right){q}={q}+\mathrm{1} \\ $$$${q}^{\mathrm{3}} +{q}^{\mathrm{2}} −{c}^{\mathrm{2}} =\mathrm{0} \\ $$$$….. \\ $$