{ ((m^2 =n+2)),((n^2 =m+2)) :} ⇒m≠n 4mn−m^3 −n^3 =?


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Question Number 155801 by cortano last updated on 05/Oct/21
${ ((m^2 =n+2)),((n^2 =m+2)) :} ⇒m≠n 4mn−m^3 −n^3 =?$
${\begin{cases} m^{2} = n + 2 \\ n^{2} = m + 2 \end{cases} \Rightarrow m \neq n$ $4 mn - m^{3} - n^{3} = ?$
	Answered by Rasheed.Sindhi last updated on 05/Oct/21
	${ ((m^2 =n+2...(i))),((n^2 =m+2...(ii))) :} ∧ m≠n 4mn−m^3 −n^3 =? (i)−(ii): m^2 −n^2 =n−m (m−n)(m+n)=−(m−n) m+n=−1 (m+n)^3 =(−1)^3 m^3 +n^3 +3mn(m+n)=−1 −m^3 −n^3 −3mn(−1)=1 3mn−m^3 −n^3 =1............A (i)×(ii): m^2 n^2 =mn+2(m+n)+4 (mn)^2 −mn−2(−1)−4=0 (mn)^2 −mn−2=0 (mn+1)(mn−2)=0 mn=−1,mn=2 A⇒3mn−m^3 −n^3 +mn=1+mn { ((4mn−m^3 −n^3 =1+(−1)=0)),((4mn−m^3 −n^3 =1+(2)=3)) :}$
	${\begin{cases} m^{2} = n + 2 . . . (i) \\ n^{2} = m + 2 . . . (i i) \end{cases} \land m \neq n$ $4 mn - m^{3} - n^{3} = ?$ $(i) - (i i) : m^{2} - n^{2} = n - m$ $(m - n) (m + n) = - (m - n)$ $m + n = - 1$ ${(m + n)}^{3} = {(- 1)}^{3}$ $m^{3} + n^{3} + 3 m n (m + n) = - 1$ $- m^{3} - n^{3} - 3 m n (- 1) = 1$ $3 m n - m^{3} - n^{3} = 1 . . . . . . . . . . . . A$ $(i) \times (i i) : m^{2} n^{2} = m n + 2 (m + n) + 4$ ${(m n)}^{2} - m n - 2 (- 1) - 4 = 0$ ${(m n)}^{2} - m n - 2 = 0$ $(m n + 1) (m n - 2) = 0$ $m n = - 1, m n = 2$ $A \Rightarrow 3 m n - m^{3} - n^{3} + m n = 1 + m n$ ${\begin{cases} 4 m n - m^{3} - n^{3} = 1 + (- 1) = 0 \\ 4 m n - m^{3} - n^{3} = 1 + (2) = 3 \end{cases}$
	Commented by cortano last updated on 05/Oct/21

	$sorry sir . not correct .$ $if 4 mn - m^{3} - n^{3} = 0 it$ $follows that m = n = 2 .$
	Commented by Rasheed.Sindhi last updated on 05/Oct/21

	$∙ 4 m n - m^{3} - n^{3} = 0 ⇏ m = n = 2$ $∙ W h a t a b o u t o t h e r a n s w e r :$ $4 m n - m^{3} - n^{3} = 1 + (2) = 3$ $I f {i t}^{'} s a l s o i n c o r r e c t, {w h a t}^{'} s r i g h t$ $a n s w e r ?$
	Commented by mr W last updated on 07/Oct/21

	$c o r t a n o s i r :$ $b o t h r a s h e e d s i r a n d i h a v e g i v e n a$ $d e t a i l e d w o r k i n g f o r t h e r e s u l t z e r o .$ $y o u s a i d t h i s r e s u l t i s w r o n g, b u t$ $y o u c a n n o t t e l l u s w h e r e i t i s w r o n g$ $i n o u r w o r k i n g . i t h i n k n o b o d y c a n t e l l$ $b e c a u s e b o t h w o r k i n g a n d r e s u l t$ $a r e c o r r e c t .$ $y o u r a n s w e r t h a t t h e a n s w e r i s n o t$ $z e r o i s c l e a r l y w r o n g . i n f a c t t o g e t$ $t h e r e s u l t w e {d o n}^{'} t n e e d t o c a l c u l a t e$ $t h e v a l u e s o f m a n d n . {i t}^{'} s o k y o u d i d .$ $y o u g o t t h e c o r r e c t v a l u e s f o r m, n :$ $n, m = \frac{- 1 \pm \sqrt{5}}{2} ✓ .$ $b u t w i t h t h e s e s v a l u e s y o u s h o u l d a n d$ $m u s t g e t 4 m m - m^{3} - n^{3} = 0 . i f n o t,$ $y o u s h o u l d b u y a n e w c a l c u l a t o r .$
	Commented by Rasheed.Sindhi last updated on 07/Oct/21
	مھرباني منھنجا سائين!
	Commented by Rasheed.Sindhi last updated on 07/Oct/21

	$S i r m r W, h o w c a n w e r e j e c t mn = 2$ $i n m y a b o v e a n w s e r ?$
	Commented by mr W last updated on 07/Oct/21

	$a n s w e r 0 i s c o r r e c t!$
	Commented by cortano last updated on 07/Oct/21

	$oo yes m \neq n \neq 2 but m = n = \frac{\pm \sqrt{5} - 1}{2}$ $then 4 mn - m^{3} - n^{3} = 0 .$ $sorry sir$
	Commented by Rasheed.Sindhi last updated on 07/Oct/21

	$B u t y o u h a v e g i v e n c o n d i t i o n m r c o r t a n o$ $t h a t m \neq n, t h e n h o w m = n = \frac{\pm \sqrt{5} - 1}{2} ?$
	Commented by Rasheed.Sindhi last updated on 07/Oct/21
	$m=n=((±(√5)−1)/2) { ((m^2 =n+2⇒((((√5)−1)/2))^2 =^(?) (((√5)−1)/2)+2)),((n^2 =m+2⇒((((√5)−1)/2))^2 =^(?) (((√5)−1)/2)+2)) :} { ((m^2 =n+2⇒((((√5)−1)/2))^2 =^(?) (((√5)−1)/2)+2)),((n^2 =m+2⇒((((√5)−1)/2))^2 =^(?) (((√5)−1)/2)+2)) :} ((5+1−2(√5))/4)=^(?) (((√5)−1+4)/2) ((3−(√5))/2)≠(((√5)+3)/2) Don′t satisfied by m=n=((±(√5)−1)/2)$
	$m = n = \frac{\pm \sqrt{5} - 1}{2}$ ${\begin{cases} m^{2} = n + 2 \Rightarrow {(\frac{\sqrt{5} - 1}{2})}^{2} \overset{?}{=} \frac{\sqrt{5} - 1}{2} + 2 \\ n^{2} = m + 2 \Rightarrow {(\frac{\sqrt{5} - 1}{2})}^{2} \overset{?}{=} \frac{\sqrt{5} - 1}{2} + 2 \end{cases}$ ${\begin{cases} m^{2} = n + 2 \Rightarrow {(\frac{\sqrt{5} - 1}{2})}^{2} \overset{?}{=} \frac{\sqrt{5} - 1}{2} + 2 \\ n^{2} = m + 2 \Rightarrow {(\frac{\sqrt{5} - 1}{2})}^{2} \overset{?}{=} \frac{\sqrt{5} - 1}{2} + 2 \end{cases}$ $\frac{5 + 1 - 2 \sqrt{5}}{4} \overset{?}{=} \frac{\sqrt{5} - 1 + 4}{2}$ $\frac{3 - \sqrt{5}}{2} \neq \frac{\sqrt{5} + 3}{2}$ ${D o n}^{'} t s a t i s f i e d b y m = n = \frac{\pm \sqrt{5} - 1}{2}$
	Commented by mr W last updated on 07/Oct/21

	$r a s h e e d s i r :$ $w i t h m + n = - 1$ $m = - (n + 1)$ $P = m n = - (1 + n) n = \frac{1}{4} - {(n + \frac{1}{2})}^{2} ⩽ \frac{1}{4}$ $t h e r e f o r e m n = 2 > \frac{1}{4} m u s t b e r e j e c t e d .$
	Commented by cortano last updated on 07/Oct/21

	Commented by cortano last updated on 07/Oct/21

	$clear the answer is not zero$
	Commented by Rasheed.Sindhi last updated on 07/Oct/21

	$THANKS - a - l o t S i r m r W!$
	Commented by cortano last updated on 07/Oct/21
	$the correct answer is 9+(√5) not zero. if 4mn−m^3 −n^3 =0 it follows that m=n=2. my solution { ((m=(((√5)−1)/2))),((n=((−(√5)−1)/2))) :}$
	$the correct answer is 9 + \sqrt{5}$ $not zero . if 4 mn - m^{3} - n^{3} = 0$ $it follows that m = n = 2 .$ $my solution {\begin{cases} m = \frac{\sqrt{5} - 1}{2} \\ n = \frac{- \sqrt{5} - 1}{2} \end{cases}$
	Commented by mr W last updated on 07/Oct/21
	$why don′t you calculate 4mn−m^3 −n^3 simply with your values: { ((m=(((√5)−1)/2))),((n=((−(√5)−1)/2))) :} here i check for you: 4mn=4×(((√5)−1)/2)×((−(√5)−1)/2) =−((√5)−1)((√5)+1) =−[((√5))^2 −1^2 ] =−(5−1) =−(4) =−4 ≠5+(√5) besides 4mn−m^3 −n^3 =0 doesn′t lead to m=n=2 only! in fact 4mn−m^3 −n^3 =0 has infinite solutions. m=n=2 is only one of infinite many! [edited]$
	$w h y {d o n}^{'} t y o u c a l c u l a t e 4 m n - m^{3} - n^{3}$ $s i m p l y w i t h y o u r v a l u e s :$ ${\begin{cases} m = \frac{\sqrt{5} - 1}{2} \\ n = \frac{- \sqrt{5} - 1}{2} \end{cases}$ $h e r e i c h e c k f o r y o u :$ $4 m n = 4 \times \frac{\sqrt{5} - 1}{2} \times \frac{- \sqrt{5} - 1}{2}$ $= - (\sqrt{5} - 1) (\sqrt{5} + 1)$ $= - [{(\sqrt{5})}^{2} - 1^{2}]$ $= - (5 - 1)$ $= - (4)$ $= - 4$ $\neq 5 + \sqrt{5}$ $b e s i d e s 4 m n - m^{3} - n^{3} = 0 {d o e s n}^{'} t$ $l e a d t o m = n = 2 o n l y! i n f a c t$ $4 m n - m^{3} - n^{3} = 0 h a s i n f i n i t e$ $s o l u t i o n s . m = n = 2 i s o n l y o n e o f$ $i n f i n i t e m a n y!$ $[e d i t e d]$
	Answered by john_santu last updated on 07/Oct/21

	$(1) m^{2} + n^{2} = m + n + 4$ $(2) m^{3} + n^{3} = (m + n) (m^{2} + n^{2} - m n)$ $= (m + n) (m + n - m n + 4)$ $(3) {(m^{2} - 2)}^{2} = m + 2$ $\Rightarrow m^{4} - 4 m^{2} - m + 2 = 0$ $\Rightarrow (m + 1) (m - 2) (m^{2} + m - 1) = 0$ $m = - 1 \land m = 2 r e j e c t e d s i n c e m \neq n$ $\Rightarrow m = \frac{- 1 \pm \sqrt{5}}{2} \land n = \frac{\mp 2 \sqrt{5} - 2}{4}$
	Commented by cortano last updated on 05/Oct/21

	$thanks$
	Answered by mr W last updated on 07/Oct/21
	$m^2 −n^2 =n−m (m+n)(m−n)=n−m ⇒m+n=−1 m^2 +n^2 =m+n+4 (m+n)^2 −2mn=m+n+4 1−2mn=−1+4 ⇒mn=−1 { m,n are roots of z^2 +z−1=0 ⇒m,n=z=((−1±(√5))/2) } (m+n)^3 =m^3 +n^3 +3mn(m+n) (−1)^3 =m^3 +n^3 +3(−1)(−1) ⇒m^3 +n^3 =−4 4mn−m^3 −n^3 =4(−1)−(−4)=0$
	$m^{2} - n^{2} = n - m$ $(m + n) (m - n) = n - m$ $\Rightarrow m + n = - 1$ $m^{2} + n^{2} = m + n + 4$ ${(m + n)}^{2} - 2 m n = m + n + 4$ $1 - 2 m n = - 1 + 4$ $\Rightarrow m n = - 1$ ${$ $m, n a r e r o o t s o f z^{2} + z - 1 = 0$ $\Rightarrow m, n = z = \frac{- 1 \pm \sqrt{5}}{2}$ $}$ ${(m + n)}^{3} = m^{3} + n^{3} + 3 m n (m + n)$ ${(- 1)}^{3} = m^{3} + n^{3} + 3 (- 1) (- 1)$ $\Rightarrow m^{3} + n^{3} = - 4$ $4 m n - m^{3} - n^{3} = 4 (- 1) - (- 4) = 0$
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