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Question Number 176636 by cortano1 last updated on 23/Sep/22

{ ((p^3 +q^3 =r^2 )),((p^3 +r^3 =q^2 )),((q^3 +r^3 =p^2 )) :} ⇒20pqr =?

$$\:\begin{cases}{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} ={r}^{\mathrm{2}} }\\{{p}^{\mathrm{3}} +{r}^{\mathrm{3}} ={q}^{\mathrm{2}} }\\{{q}^{\mathrm{3}} +{r}^{\mathrm{3}} ={p}^{\mathrm{2}} }\end{cases} \\ $$$$\:\Rightarrow\mathrm{20}{pqr}\:=? \\ $$

Commented by Frix last updated on 23/Sep/22

easy to see p=q=r=0∨(1/2) I think for p, q, r ∈C there are solutions with p≠q∧q=r (or rotations of this) and p≠q∧q≠r

$$\mathrm{easy}\:\mathrm{to}\:\mathrm{see}\:{p}={q}={r}=\mathrm{0}\vee\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{for}\:{p},\:{q},\:{r}\:\in\mathbb{C}\:\mathrm{there}\:\mathrm{are}\:\mathrm{solutions} \\ $$$$\mathrm{with}\:{p}\neq{q}\wedge{q}={r}\:\left(\mathrm{or}\:\mathrm{rotations}\:\mathrm{of}\:\mathrm{this}\right) \\ $$$$\:\mathrm{and}\:{p}\neq{q}\wedge{q}\neq{r} \\ $$

Commented by Frix last updated on 23/Sep/22

one possibility is pqr=(1/2)

$$\mathrm{one}\:\mathrm{possibility}\:\mathrm{is}\:{pqr}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Answered by BaliramKumar last updated on 23/Sep/22

p = q = r = (1/2) 20pqr = 20×(1/2)×(1/2)×(1/2) = (5/2)

$${p}\:=\:{q}\:=\:{r}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{20}{pqr}\:=\:\mathrm{20}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{5}}{\mathrm{2}} \\ $$

Commented by mr W last updated on 23/Sep/22

i think there are other solutions. example: p=q=r=0 is also ok.

$${i}\:{think}\:{there}\:{are}\:{other}\:{solutions}. \\ $$$${example}:\:{p}={q}={r}=\mathrm{0}\:{is}\:{also}\:{ok}. \\ $$

Answered by Rasheed.Sindhi last updated on 24/Sep/22

{ ((p^3 +q^3 =r^2 ...(i))),((p^3 +r^3 =q^2 ...(ii))),((q^3 +r^3 =p^2 ...(iii))) :} ; 20pqr =? (ii)⇒p^3 +(p^3 +q^3 )r=q^2 (iii)⇒q^3 +(p^3 +q^3 )r=p^2 r=((q^2 −p^3 )/(p^3 +q^3 ))=((p^2 −q^3 )/(p^3 +q^3 ))=±(√(p^3 +q^3 )) { ((q^2 −p^3 =p^2 −q^3 .....(A))),((p^2 −q^3 =±(p^3 +q^3 )^(3/2) ....(B))),((q^2 −p^3 =±(p^3 +q^3 )^(3/2) ....(C))) :} (A): q^2 −p^3 =p^2 −q^3 p^2 −q^2 +p^3 −q^3 =0 (p−q)(p+q)+(p−q)(p^2 +pq+q^2 )=0 (p−q)(p^2 +pq+q^2 +p+q)=0 p=q ∣ p^2 +pq+q^2 +p+q=0^★ In similar way q=r or p=q=r (i)⇒r^3 +r^3 =r^2 ⇒2r=1⇒r=(1/2) or p=q=r=(1/2) 20pqr=20(1/2)(1/2)(1/2)=5/2 p^2 +pq+q^2 +p+q=0 (p+q)^2 +(p+q)−pq=0 p+q=((−1±(√(1+4pq)))/2) (B): p^2 −q^3 =±(p^3 +q^3 )^(3/2) (p^2 −q^3 )^2 =(p^3 +q^3 )^3 p^4 −2p^2 q^3 +q^6 =p^9 +q^9 +3(pq)^3 (p^3 +q^3 ) (C): q^2 −p^3 =±(p^3 +q^3 )^(3/2) Continue...

$$\:\begin{cases}{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} ={r}^{\mathrm{2}} ...\left({i}\right)}\\{{p}^{\mathrm{3}} +{r}^{\mathrm{3}} ={q}^{\mathrm{2}} ...\left({ii}\right)}\\{{q}^{\mathrm{3}} +{r}^{\mathrm{3}} ={p}^{\mathrm{2}} ...\left({iii}\right)}\end{cases}\:;\:\:\:\:\mathrm{20}{pqr}\:=? \\ $$$$\left({ii}\right)\Rightarrow{p}^{\mathrm{3}} +\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right){r}={q}^{\mathrm{2}} \\ $$$$\left({iii}\right)\Rightarrow{q}^{\mathrm{3}} +\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right){r}={p}^{\mathrm{2}} \\ $$$${r}=\frac{{q}^{\mathrm{2}} −{p}^{\mathrm{3}} }{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} }=\frac{{p}^{\mathrm{2}} −{q}^{\mathrm{3}} }{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} }=\pm\sqrt{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} }\: \\ $$$$\begin{cases}{{q}^{\mathrm{2}} −{p}^{\mathrm{3}} ={p}^{\mathrm{2}} −{q}^{\mathrm{3}} .....\left({A}\right)}\\{{p}^{\mathrm{2}} −{q}^{\mathrm{3}} =\pm\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)^{\mathrm{3}/\mathrm{2}} ....\left({B}\right)}\\{{q}^{\mathrm{2}} −{p}^{\mathrm{3}} =\pm\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)^{\mathrm{3}/\mathrm{2}} ....\left({C}\right)}\end{cases} \\ $$$$\left({A}\right): \\ $$$${q}^{\mathrm{2}} −{p}^{\mathrm{3}} ={p}^{\mathrm{2}} −{q}^{\mathrm{3}} \\ $$$${p}^{\mathrm{2}} −{q}^{\mathrm{2}} +{p}^{\mathrm{3}} −{q}^{\mathrm{3}} =\mathrm{0} \\ $$$$\left({p}−{q}\right)\left({p}+{q}\right)+\left({p}−{q}\right)\left({p}^{\mathrm{2}} +{pq}+{q}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\left({p}−{q}\right)\left({p}^{\mathrm{2}} +{pq}+{q}^{\mathrm{2}} +{p}+{q}\right)=\mathrm{0} \\ $$$${p}={q}\:\mid\:{p}^{\mathrm{2}} +{pq}+{q}^{\mathrm{2}} +{p}+{q}=\mathrm{0}^{\bigstar} \\ $$$${In}\:{similar}\:{way}\:{q}={r} \\ $$$${or}\:\:{p}={q}={r} \\ $$$$\left({i}\right)\Rightarrow{r}^{\mathrm{3}} +{r}^{\mathrm{3}} ={r}^{\mathrm{2}} \Rightarrow\mathrm{2}{r}=\mathrm{1}\Rightarrow{r}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${or}\:\:{p}={q}={r}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{20}{pqr}=\mathrm{20}\left(\mathrm{1}/\mathrm{2}\right)\left(\mathrm{1}/\mathrm{2}\right)\left(\mathrm{1}/\mathrm{2}\right)=\mathrm{5}/\mathrm{2} \\ $$$$ \\ $$$${p}^{\mathrm{2}} +{pq}+{q}^{\mathrm{2}} +{p}+{q}=\mathrm{0} \\ $$$$\:\:\:\:\:\left({p}+{q}\right)^{\mathrm{2}} +\left({p}+{q}\right)−{pq}=\mathrm{0} \\ $$$$\:\:\:\:{p}+{q}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}{pq}}}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$\left({B}\right): \\ $$$${p}^{\mathrm{2}} −{q}^{\mathrm{3}} =\pm\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)^{\mathrm{3}/\mathrm{2}} \\ $$$$\left({p}^{\mathrm{2}} −{q}^{\mathrm{3}} \right)^{\mathrm{2}} =\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)^{\mathrm{3}} \\ $$$${p}^{\mathrm{4}} −\mathrm{2}{p}^{\mathrm{2}} {q}^{\mathrm{3}} +{q}^{\mathrm{6}} ={p}^{\mathrm{9}} +{q}^{\mathrm{9}} +\mathrm{3}\left({pq}\right)^{\mathrm{3}} \left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right) \\ $$$$\left({C}\right): \\ $$$${q}^{\mathrm{2}} −{p}^{\mathrm{3}} =\pm\left({p}^{\mathrm{3}} +{q}^{\mathrm{3}} \right)^{\mathrm{3}/\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{Continue}... \\ $$

Commented by Frix last updated on 24/Sep/22

it′s easier this way: q=αp∧r=βp { (((1+α^3 )p^3 =β^2 p^2 )),(((1+β^3 )p^3 =α^2 p^2 )),(((α^3 +β^3 )p^3 =p^2 )) :} first obvious solution p=0 ⇒ p=q=r=0 { ((p=(β^2 /(1+α^3 )))),((p=(α^2 /(1+β^3 )))),((p=(1/(α^3 +β^3 )))) :} second obvious solution α=β=1 ⇒ p=q=r=(1/2) now we have to solve (1/(α^3 +β^3 ))=(α^2 /(1+β^3 ))=(β^2 /(1+α^3 )) which is hard I think we get 19 pairs (α, β) including (1, 1) >and of course because we can exchange α and β 9 of the remaining 18 are unique if we set β=1 ⇒ α=1 and 2 pairs of complex solutions which I can only give approximately at the moment α=−.975564±.528237 α=.475564±1.18273 the equation is α^4 +α^3 +α^2 +2α+2=0 and has no useable exact solutions

$$\mathrm{it}'\mathrm{s}\:\mathrm{easier}\:\mathrm{this}\:\mathrm{way}: \\ $$$${q}=\alpha{p}\wedge{r}=\beta{p} \\ $$$$\begin{cases}{\left(\mathrm{1}+\alpha^{\mathrm{3}} \right){p}^{\mathrm{3}} =\beta^{\mathrm{2}} {p}^{\mathrm{2}} }\\{\left(\mathrm{1}+\beta^{\mathrm{3}} \right){p}^{\mathrm{3}} =\alpha^{\mathrm{2}} {p}^{\mathrm{2}} }\\{\left(\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} \right){p}^{\mathrm{3}} ={p}^{\mathrm{2}} }\end{cases} \\ $$$$\mathrm{first}\:\mathrm{obvious}\:\mathrm{solution}\:{p}=\mathrm{0}\:\Rightarrow\:{p}={q}={r}=\mathrm{0} \\ $$$$\begin{cases}{{p}=\frac{\beta^{\mathrm{2}} }{\mathrm{1}+\alpha^{\mathrm{3}} }}\\{{p}=\frac{\alpha^{\mathrm{2}} }{\mathrm{1}+\beta^{\mathrm{3}} }}\\{{p}=\frac{\mathrm{1}}{\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} }}\end{cases} \\ $$$$\mathrm{second}\:\mathrm{obvious}\:\mathrm{solution}\:\alpha=\beta=\mathrm{1}\:\Rightarrow\:{p}={q}={r}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{solve} \\ $$$$\frac{\mathrm{1}}{\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} }=\frac{\alpha^{\mathrm{2}} }{\mathrm{1}+\beta^{\mathrm{3}} }=\frac{\beta^{\mathrm{2}} }{\mathrm{1}+\alpha^{\mathrm{3}} } \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{hard} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{we}\:\mathrm{get}\:\mathrm{19}\:\mathrm{pairs}\:\left(\alpha,\:\beta\right)\:\mathrm{including}\:\left(\mathrm{1},\:\mathrm{1}\right) \\ $$$$\:>\mathrm{and}\:\mathrm{of}\:\mathrm{course}\:\mathrm{because}\:\mathrm{we}\:\mathrm{can}\:\mathrm{exchange} \\ $$$$\alpha\:\mathrm{and}\:\beta\:\mathrm{9}\:\mathrm{of}\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{18}\:\mathrm{are}\:\mathrm{unique} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{set}\:\beta=\mathrm{1}\:\Rightarrow\:\alpha=\mathrm{1}\:\mathrm{and}\:\mathrm{2}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{complex} \\ $$$$\mathrm{solutions}\:\mathrm{which}\:\mathrm{I}\:\mathrm{can}\:\mathrm{only}\:\mathrm{give}\:\mathrm{approximately}\:\mathrm{at} \\ $$$$\mathrm{the}\:\mathrm{moment} \\ $$$$\alpha=−.\mathrm{975564}\pm.\mathrm{528237} \\ $$$$\alpha=.\mathrm{475564}\pm\mathrm{1}.\mathrm{18273} \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{is}\:\alpha^{\mathrm{4}} +\alpha^{\mathrm{3}} +\alpha^{\mathrm{2}} +\mathrm{2}\alpha+\mathrm{2}=\mathrm{0}\:\mathrm{and} \\ $$$$\mathrm{has}\:\mathrm{no}\:\mathrm{useable}\:\mathrm{exact}\:\mathrm{solutions} \\ $$

Commented by Rasheed.Sindhi last updated on 24/Sep/22

Yes sir, you′re right! ThanX a lot!

$$\mathrm{Yes}\:\mathrm{sir},\:\mathrm{you}'\mathrm{re}\:\mathrm{right}!\: \\ $$$$\mathcal{T}{han}\mathcal{X}\:{a}\:{lot}! \\ $$

Commented by Frix last updated on 24/Sep/22

I get (α, β)= (1, β=α) (.475564±1.18273i,1) (.384646±.923064i, −.500000∓.333321i) [not sure if −(1/2)] (.292652±.727828i, β=α) (−.792652±.429196i, β=α) (−.975564±.528237i, 1) (−1.38465±.923064i, β=conj(α))

$$\mathrm{I}\:\mathrm{get}\:\left(\alpha,\:\beta\right)= \\ $$$$\left(\mathrm{1},\:\beta=\alpha\right) \\ $$$$\left(.\mathrm{475564}\pm\mathrm{1}.\mathrm{18273i},\mathrm{1}\right) \\ $$$$\left(.\mathrm{384646}\pm.\mathrm{923064i},\:−.\mathrm{500000}\mp.\mathrm{333321i}\right)\:\left[\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:−\frac{\mathrm{1}}{\mathrm{2}}\right] \\ $$$$\left(.\mathrm{292652}\pm.\mathrm{727828i},\:\beta=\alpha\right) \\ $$$$\left(−.\mathrm{792652}\pm.\mathrm{429196i},\:\beta=\alpha\right) \\ $$$$\left(−.\mathrm{975564}\pm.\mathrm{528237i},\:\mathrm{1}\right) \\ $$$$\left(−\mathrm{1}.\mathrm{38465}\pm.\mathrm{923064i},\:\beta=\mathrm{conj}\left(\alpha\right)\right) \\ $$

Commented by Tawa11 last updated on 25/Sep/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

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