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r-q-1-pq-q-1-c-2-p-qr-2-help-find-p-q-r-




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

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