p-3-q-3-r-2-p-3-r-3-q-2-q-3-r-3-p-2-20pqr-

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

{\begin{cases} p^{3} + q^{3} = r^{2} \\ p^{3} + r^{3} = q^{2} \\ q^{3} + r^{3} = p^{2} \end{cases}

\Rightarrow 20 p q r = ?

Commented by Frix last updated on 23/Sep/22

easy to see p = q = r = 0 \lor \frac{1}{2}

I think for p, q, r \in C there are solutions

with p \neq q \land q = r (or rotations of this)

and p \neq q \land q \neq r

Commented by Frix last updated on 23/Sep/22

one possibility is p q r = \frac{1}{2}

Answered by BaliramKumar last updated on 23/Sep/22

p = q = r = \frac{1}{2}

20 p q r = 20 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{5}{2}

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

i t h i n k t h e r e a r e o t h e r s o l u t i o n s .

e x a m p l e : p = q = r = 0 i s a l s o o k .

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

{\begin{cases} p^{3} + q^{3} = r^{2} \dots (i) \\ p^{3} + r^{3} = q^{2} \dots (i i) \\ q^{3} + r^{3} = p^{2} \dots (i i i) \end{cases}; 20 p q r = ?

(i i) \Rightarrow p^{3} + (p^{3} + q^{3}) r = q^{2}

(i i i) \Rightarrow q^{3} + (p^{3} + q^{3}) r = p^{2}

r = \frac{q^{2} - p^{3}}{p^{3} + q^{3}} = \frac{p^{2} - q^{3}}{p^{3} + q^{3}} = \pm \sqrt{p^{3} + q^{3}}

{\begin{cases} q^{2} - p^{3} = p^{2} - q^{3} \dots . . (A) \\ p^{2} - q^{3} = \pm {(p^{3} + q^{3})}^{3 / 2} \dots . (B) \\ q^{2} - p^{3} = \pm {(p^{3} + q^{3})}^{3 / 2} \dots . (C) \end{cases}

(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} + p q + q^{2}) = 0

(p - q) (p^{2} + p q + q^{2} + p + q) = 0

p = q ∣ p^{2} + p q + q^{2} + p + q = 0^{★}

I n s i m i l a r w a y q = r

o r p = q = r

(i) \Rightarrow r^{3} + r^{3} = r^{2} \Rightarrow 2 r = 1 \Rightarrow r = \frac{1}{2}

o r p = q = r = \frac{1}{2}

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

p^{2} + p q + q^{2} + p + q = 0

{(p + q)}^{2} + (p + q) - p q = 0

p + q = \frac{- 1 \pm \sqrt{1 + 4 p q}}{2}

(B) :

p^{2} - q^{3} = \pm {(p^{3} + q^{3})}^{3 / 2}

{(p^{2} - q^{3})}^{2} = {(p^{3} + q^{3})}^{3}

p^{4} - 2 p^{2} q^{3} + q^{6} = p^{9} + q^{9} + 3 {(p q)}^{3} (p^{3} + q^{3})

(C) :

q^{2} - p^{3} = \pm {(p^{3} + q^{3})}^{3 / 2}

Continue \dots

Commented by Frix last updated on 24/Sep/22

{it}^{'} s easier this way :

q = α p \land r = β p

{\begin{cases} (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} \end{cases}

first obvious solution p = 0 \Rightarrow p = q = r = 0

{\begin{cases} p = \frac{β^{2}}{1 + α^{3}} \\ p = \frac{α^{2}}{1 + β^{3}} \\ p = \frac{1}{α^{3} + β^{3}} \end{cases}

second obvious solution α = β = 1 \Rightarrow p = q = r = \frac{1}{2}

now we have to solve

\frac{1}{α^{3} + β^{3}} = \frac{α^{2}}{1 + β^{3}} = \frac{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 \Rightarrow α = 1 and 2 pairs of complex

solutions which I can only give approximately at

the moment

α = - . 975564 \pm . 528237

α = . 475564 \pm 1.18273

the equation is α^{4} + α^{3} + α^{2} + 2 α + 2 = 0 and

has no useable exact solutions

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

Yes sir, {you}^{'} re right!

T h a n X a l o t!

Commented by Frix last updated on 24/Sep/22

I get (α, β) =

(1, β = α)

(. 475564 \pm 1 . 18273 i, 1)

(. 384646 \pm . 923064 i, - . 500000 \mp . 333321 i) [not sure if - \frac{1}{2}]

(. 292652 \pm . 727828 i, β = α)

(- . 792652 \pm . 429196 i, β = α)

(- . 975564 \pm . 528237 i, 1)

(- 1.38465 \pm . 923064 i, β = conj (α))

Commented by Tawa11 last updated on 25/Sep/22

Great sir

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