x^3 +y^3 =5 x^2 +y^2 =3


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Question Number 36692 by math2018 last updated on 04/Jun/18

$x^{3} + y^{3} = 5$ $x^{2} + y^{2} = 3$
Commented by behi83417@gmail.com last updated on 04/Jun/18
$x+y=s,xy=t (x+y)(x^2 −xy+y^2 )=5⇒s(3−t)=5 x^2 +y^2 =(x+y)^2 −2xy=s^2 −2t=3 ⇒ { ((3s−st=5)),((s^2 −2t=3)) :}⇒ { ((−6s+2st=−10)),((s^3 −2st=3s)) :}⇒ s^3 −9s+10=0⇒(s−2)(s^2 +2s−5)=0 ⇒s=2,−1±(√6) 1)s=2⇒6−2t=5⇒t=(1/2) ⇒ { ((x+y=2)),((xy=(1/2))) :}⇒z^2 −2z+(1/2)=0⇒z=1±((√2)/2) ⇒(x,y)=(1+((√2)/2),1−((√2)/2)),(1+((√2)/2),1−((√2)/2)). 2)s=−1+(√6)⇒t=3−(5/s)=3−(5/((√6)−1))= =3−((√6)+1)=2−(√6) ⇒ { ((x+y=(√6)−1)),((xy=2−(√6))) :}⇒z^2 −((√6)−1)z+(2−(√6))=0 ⇒z=((((√6)−1)±(√(((√6)−1)^2 −4(2−(√6)))))/2)= =((((√6)−1)±(√(6+1−2(√6)−8+4(√6))))/2)⇒ ⇒ { ((x=(((√6)+(√(2(√6)−1))−1)/2))),((y=(((√6)−(√(2(√6)−1))−1)/2))) :} 3)s=−1−(√6)⇒t=3−(5/s)=3+(5/((√6)+1))= =3+(√6)−1=2+(√6) ⇒ { ((x+y=−1−(√6))),((xy=2+(√6))) :}⇒z^2 +((√6)+1)z+(2+(√6))=0 ⇒z=((−((√6)+1)±(√(((√6)+1)^2 −4(2+(√6)))))/2)= =((−((√6)+1)±(√(−1−2(√6))))/2) ⇒ { ((x=((−((√6)+1)+i(√(2(√6)+1)))/2))),((y=((−((√6)+1)−i(√(2(√6)+1)))/2))) :} .■$
$x + y = s, x y = t$ $(x + y) (x^{2} - x y + y^{2}) = 5 \Rightarrow s (3 - t) = 5$ $x^{2} + y^{2} = {(x + y)}^{2} - 2 x y = s^{2} - 2 t = 3$ $\Rightarrow {\begin{cases} 3 s - s t = 5 \\ s^{2} - 2 t = 3 \end{cases} \Rightarrow {\begin{cases} - 6 s + 2 s t = - 10 \\ s^{3} - 2 s t = 3 s \end{cases} \Rightarrow$ $s^{3} - 9 s + 10 = 0 \Rightarrow (s - 2) (s^{2} + 2 s - 5) = 0$ $\Rightarrow s = 2, - 1 \pm \sqrt{6}$ $1) s = 2 \Rightarrow 6 - 2 t = 5 \Rightarrow t = \frac{1}{2}$ $\Rightarrow {\begin{cases} x + y = 2 \\ x y = \frac{1}{2} \end{cases} \Rightarrow z^{2} - 2 z + \frac{1}{2} = 0 \Rightarrow z = 1 \pm \frac{\sqrt{2}}{2}$ $\Rightarrow (x, y) = (1 + \frac{\sqrt{2}}{2}, 1 - \frac{\sqrt{2}}{2}), (1 + \frac{\sqrt{2}}{2}, 1 - \frac{\sqrt{2}}{2}) .$ $2) s = - 1 + \sqrt{6} \Rightarrow t = 3 - \frac{5}{s} = 3 - \frac{5}{\sqrt{6} - 1} =$ $= 3 - (\sqrt{6} + 1) = 2 - \sqrt{6}$ $\Rightarrow {\begin{cases} x + y = \sqrt{6} - 1 \\ x y = 2 - \sqrt{6} \end{cases} \Rightarrow z^{2} - (\sqrt{6} - 1) z + (2 - \sqrt{6}) = 0$ $\Rightarrow z = \frac{(\sqrt{6} - 1) \pm \sqrt{{(\sqrt{6} - 1)}^{2} - 4 (2 - \sqrt{6})}}{2} =$ $= \frac{(\sqrt{6} - 1) \pm \sqrt{6 + 1 - 2 \sqrt{6} - 8 + 4 \sqrt{6}}}{2} \Rightarrow$ $\Rightarrow {\begin{cases} x = \frac{\sqrt{6} + \sqrt{2 \sqrt{6} - 1} - 1}{2} \\ y = \frac{\sqrt{6} - \sqrt{2 \sqrt{6} - 1} - 1}{2} \end{cases}$ $3) s = - 1 - \sqrt{6} \Rightarrow t = 3 - \frac{5}{s} = 3 + \frac{5}{\sqrt{6} + 1} =$ $= 3 + \sqrt{6} - 1 = 2 + \sqrt{6}$ $\Rightarrow {\begin{cases} x + y = - 1 - \sqrt{6} \\ x y = 2 + \sqrt{6} \end{cases} \Rightarrow z^{2} + (\sqrt{6} + 1) z + (2 + \sqrt{6}) = 0$ $\Rightarrow z = \frac{- (\sqrt{6} + 1) \pm \sqrt{{(\sqrt{6} + 1)}^{2} - 4 (2 + \sqrt{6}})}{2} =$ $= \frac{- (\sqrt{6} + 1) \pm \sqrt{- 1 - 2 \sqrt{6}}}{2}$ $\Rightarrow {\begin{cases} x = \frac{- (\sqrt{6} + 1) + i \sqrt{2 \sqrt{6} + 1}}{2} \\ y = \frac{- (\sqrt{6} + 1) - i \sqrt{2 \sqrt{6} + 1}}{2} \end{cases} . ◼$
	Answered by Joel579 last updated on 04/Jun/18
	$a = x + y b = xy a^2 − 2b = 3 → b = ((a^2 − 3)/2) a(3 − b) = 5 a(((9 − a^2 )/2)) = 5 9a − a^3 = 10 a^3 − 9a + 10 = 0 (a − 2)(a^2 + 2a − 5) = 0 • a − 2 = 0 → a = 2 → b = (1/2) { ((x + y = 2)),((xy = (1/2))) :} • a^2 + 2a − 5 = 0 a = −1 + (√6) → b = 2 − (√6) { ((x + y = −1 + (√6))),((xy = 2 − (√6))) :} a = −1 − (√6) → b = 2 + (√6) { ((x + y = −1 − (√6))),((xy = 2 + (√6))) :}$
	$a = x + y$ $b = x y$ $a^{2} - 2 b = 3 \to b = \frac{a^{2} - 3}{2}$ $a (3 - b) = 5$ $a (\frac{9 - a^{2}}{2}) = 5$ $9 a - a^{3} = 10$ $a^{3} - 9 a + 10 = 0$ $(a - 2) (a^{2} + 2 a - 5) = 0$ $∙ a - 2 = 0 \to a = 2 \to b = \frac{1}{2}$ ${\begin{cases} x + y = 2 \\ x y = \frac{1}{2} \end{cases}$ $∙ a^{2} + 2 a - 5 = 0$ $a = - 1 + \sqrt{6} \to b = 2 - \sqrt{6}$ ${\begin{cases} x + y = - 1 + \sqrt{6} \\ x y = 2 - \sqrt{6} \end{cases}$ $a = - 1 - \sqrt{6} \to b = 2 + \sqrt{6}$ ${\begin{cases} x + y = - 1 - \sqrt{6} \\ x y = 2 + \sqrt{6} \end{cases}$
	Commented by Joel579 last updated on 04/Jun/18

	$Now {it}^{'} s easier to find (x, y)$
	Commented by math2018 last updated on 04/Jun/18

	$T h a n k s S i r$
	Answered by MJS last updated on 04/Jun/18

	$looking for solutions of the form (because$ $I found no integer solutions and otherwise$ $we {can}^{'} t exactly solve) :$ $x = a \sqrt{b} + c \sqrt{d} \land y = a \sqrt{b} - c \sqrt{d} \Rightarrow$ $\Rightarrow x = p (1 + q) \land y = p (1 - q)$ $(1) p^{2} {(1 + q)}^{2} + p^{2} {(1 - q)}^{2} = 3$ $(2) p^{3} {(1 + q)}^{3} + p^{3} {(1 - q)}^{3} = 5$ $(1) p^{2} (2 q^{2} + 2) = 3 \Rightarrow q^{2} = \frac{3}{2 p^{2}} - 1$ $(2) p^{3} (6 q^{2} + 2) = 5 \Rightarrow q^{2} = \frac{5}{6 p^{3}} - \frac{1}{3}$ $\frac{3}{2 p^{2}} - 1 = \frac{5}{6 p^{3}} - \frac{1}{3} \Rightarrow p^{3} - \frac{9}{4} p + \frac{5}{4} = 0 \Rightarrow$ $\Rightarrow (p - 1) (p^{2} + p - \frac{5}{4}) = 0$ $p_{1} = 1 \Rightarrow q_{1} = \pm \frac{\sqrt{2}}{2}$ $p_{2} = - \frac{1}{2} - \frac{\sqrt{6}}{2} \Rightarrow q_{2} = \pm \frac{\sqrt{- 17 + 12 \sqrt{6}}}{5} i$ $p_{3} = - \frac{1}{2} + \frac{\sqrt{6}}{2} \Rightarrow q_{3} = \pm \frac{\sqrt{17 + 12 \sqrt{6}}}{5}$ $x_{1} = 1 \pm \frac{\sqrt{2}}{2}; y_{1} = 1 \mp \frac{\sqrt{2}}{2}$ $x_{2} = - \frac{1}{2} - \frac{\sqrt{6}}{2} \pm \frac{\sqrt{1 + 2 \sqrt{6}}}{2} i;$ $y_{2} = - \frac{1}{2} - \frac{\sqrt{6}}{2} \mp \frac{\sqrt{1 + 2 \sqrt{6}}}{2} i$ $x_{3} = - \frac{1}{2} + \sqrt{6} \pm \frac{\sqrt{- 1 + 2 \sqrt{6}}}{2};$ $y_{3} = - \frac{1}{2} + \sqrt{6} \mp \frac{\sqrt{- 1 + 2 \sqrt{6}}}{2}$
	Commented by math2018 last updated on 04/Jun/18

	$T h a n k s f o r y o u r w o r k .$
	Commented by MJS last updated on 04/Jun/18

	${you}^{'} re welcome$
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