if x^2 +y^2 =p, x^3 +y^3 =q, find x^n +y^n in terms of p, q and n. (n≥4)


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Question Number 75778 by mr W last updated on 16/Dec/19

$i f x^{2} + y^{2} = p, x^{3} + y^{3} = q,$ $f i n d x^{n} + y^{n} i n t e r m s o f p, q a n d n .$ $(n ⩾ 4)$
	Answered by mind is power last updated on 17/Dec/19
	$let e_1 =x+y,e_2 =xy e_k =0,∀k≥3 p_k =x^k +y^k ⇒ { ((2e_2 =e_1 .p_1 −p_2 )),((3e_3 =0=e_2 p_1 −e_1 p_2 +p_3 )) :}⇔ { ((2e_2 =p_1 ^2 −p)),((0=e_2 p_1 −p_1 p+q)) :} ⇒0=p_1 (((p_1 ^2 −p)/2)−p)+q⇔p_1 ^3 −3pp_1 +2q=0 p_1 Root of X^3 −3pX+2q cardan p_1 know ⇒e_2 know { ((x+y=p_1 )),((xy=e_2 )) :} ⇒x,y Root of X^2 −p_1 X+e_2 lets say a,b X^n +Y^n =a^n +b^n$
	$let e_{1} = x + y, e_{2} = xy$ $e_{k} = 0, \forall k ⩾ 3$ $p_{k} = x^{k} + y^{k}$ $\Rightarrow {\begin{cases} {2 e}_{2} = e_{1} . p_{1} - p_{2} \\ {3 e}_{3} = 0 = e_{2} p_{1} - e_{1} p_{2} + p_{3} \end{cases} \Leftrightarrow$ ${\begin{cases} {2 e}_{2} = p_{1}^{2} - p \\ 0 = e_{2} p_{1} - p_{1} p + q \end{cases}$ $\Rightarrow 0 = p_{1} (\frac{p_{1}^{2} - p}{2} - p) + q \Leftrightarrow p_{1}^{3} - {3 pp}_{1} + 2 q = 0$ $p_{1} Root of$ $X^{3} - 3 pX + 2 q cardan$ $p_{1} know \Rightarrow e_{2} know$ ${\begin{cases} x + y = p_{1} \\ xy = e_{2} \end{cases}$ $\Rightarrow x, y Root of X^{2} - p_{1} X + e_{2} lets say a, b$ $X^{n} + Y^{n} = a^{n} + b^{n}$
	Commented by mr W last updated on 18/Dec/19

	$t h a n k y o u s i r!$
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