How to solve the above question from quadratic equations?

Question

If $a$ and $b$ are the roots of the equation $x^{2}$$-$7$x$$-$1$=$0, then find the value of $($$a^{21}$$+$$b^{21}$$+$$a^{17}$$+$$b^{17}$$)$$/$$($$a^{19}$$+$$b^{19}$$)$$?$ I am trying this question by taking $a^{17}$ common from $($$a^{21}$$+$$a^{17}$$)$ and $b^{17}$ common from$($$b^{21}$$+$$b^{17}$$)$. So, finally the expression is becoming [$a^{17}$$($$a^{4}$$+$1$)$+$b^{17}$$($$b^{4}$$+$1$)$]$/$$($$a^{19}$$+$$b^{19}$$)$. But I can't proceed further. Can it be solved by finding the roots $a$ and $b$? Please help me out.

Answer 1

If I were to write $$ a^{21} + b^{21} = (a^{19} + b^{19})(a^2 + b^2 - s) $$ where the s just stands for “something I don't know yet”, then I can do that first step of polynomial division to get $$ s= \frac{a^{19}b^{2} + b^{19}a^2}{a^{19}+b^{19}}$$but at first this doesn't look like it simplified anything because we still have a total exponent of 21 up-top.

But, since I know that $$ (x-a)(x-b) = x^2 -7x -1,\\ a+b=7,~~~~ab={-1}, $$ this is actually $$ s= (-1)^2\frac{a^{17} + b^{17}}{a^{19}+b^{19}}$$ thus$$ {a^{21} + b^{21}+a^{17}+b^{17}\over a^{19} + b^{19}}=a^2 + b^2 $$ is true in this particular case even though it is not true in general.

Then in this particular case,

$$49 = (a+b)^2 = a^2+b^2 + 2ab$$ and the rest is simple addition.

Answer 2

Let $u_n = a^n+b^n$. Then by Newton's theorem, $u_n =7u_{n-1}+u_{n-2}$ or $u_n-u_{n-2}=7u_{n-1}$

Let $A=\dfrac{u_{21}+u_{17}}{u_{19}}$.

Then $A-2 = \dfrac{\left(u_{21}-u_{19}\right) -\left(u_{19}-u_{17}\right)}{u_{19}}$

$=\dfrac{7u_{20} - 7 u_{18}}{u_{19}}=\dfrac{49u_{19}}{u_{19}}=49$

$\Rightarrow A=51$