No use of any identities

By Vieta's formulas $a_1=1$ , by Newton's sums $a_2-a_1-2=0$ , i.e. $a_2=3$ .

By Newton's sums : $a_k=a_{k-1}+a_{k-2}$ for $k\ge 3$ .

With this recurrence relation find $a_{20},a_{22}$ .

In fact, since $a_0=2$ , we have $a_k=L_k$ for all $k\ge 0$ , where $(L_n)_{n\ge 0}$ is the sequence of Lucas numbers . So you can just see a list of those numbers, e.g. here .

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Alternatively (more elementary):

By Vieta's formulas $a_1=1, \alpha\beta=-1$ .

$a_2=a_1^2-2\alpha\beta=3$ .

$a_k=a_{k-1}a_1-\alpha\beta a_{k-2}$ , i.e.

$a_k=a_{k-1}+a_{k-2}$ .

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Another way:

$a_4=a_2^2-2\alpha^2\beta^2$

$a_8=a_4^2-2\alpha^4\beta^4$

$a_{16}=a_8^2-2\alpha^8\beta^8$

$a_{20}=a_{16}a_{4}-\alpha^4\beta^4(a_8a_4-\alpha^4\beta^4a_4)$

$a_{22}=a_{20}a_2-\alpha^2\beta^2(a_{16}a_2-\alpha^2\beta^2(a_8(a_4a_2-\alpha^2\beta^2a_2)-\alpha^6\beta^6a_2))$

Or $a_{22}=a_{16}(a_4a_2-\alpha^2\beta^2a_2)-\alpha^6\beta^6(a_8a_2-\alpha^2\beta^2(a_4a_2-\alpha^2\beta^2a_2))$

The equation is $x^2-x-1=0$

Since $\beta$ is a root,

${\beta}^2-{\beta}-1=0$ .

Multiplying both sides by ${\beta}^{2n+2}$

$\Rightarrow {\beta}^{2n+4}-{\beta}^{2n+3}-{\beta}^{2n+2}=0$ ................................. $(i)$

and multiplying both sides by ${\beta}^{2n+1}$

$\Rightarrow {\beta}^{2n+3}-{\beta}^{2n+2}-{\beta}^{2n+1}=0$ ................................. $(ii)$

and multiplying both sides by ${\beta}^{2n}$

$\Rightarrow {\beta}^{2n+2}-{\beta}^{2n+1}-{\beta}^{2n}=0$ ................................. $(iii)$

Now $(i)+(ii)-(iii)$ we get;

${\beta}^{2n+4}-3{\beta}^{2n+2}+{\beta}^{2n}=0$ ............................. $(iv)$

Similarly;

${\alpha}^{2n+4}-3{\alpha}^{2n+2}+{\alpha}^{2n}=0$ ......................... $(v)$

Now $(iv)+(v)$

$({\beta}^{2n+4}+{\alpha}^{2n+4})-3({\beta}^{2n+2}+{\alpha}^{2n+2})+({\beta}^{2n}+{\alpha}^{2n})=0$ .

$\Rightarrow a_{2n+4}-3a_{2n+2}+a_{2n}=0$ .

$a_{2}={\alpha}^2+{\beta}^2=({\alpha}+{\beta})^2-2{\alpha}{\beta}=1+2=3$

$a_{4}={\alpha}^4+{\beta}^4=({\alpha}^2+{\beta}^2)^2-2{{\alpha}^2}{{\beta}^2}=9-2=7$ .

Now using the recurrence relation derived we can easily calculate

$a_{20}=15127 \quad a_{22}=39603$

$\Rightarrow a_{20}+a_{22}=\boxed{54730}$

(there is another way of deriving the recurrence relation that is by forming a quadratic equation with roots as ${\alpha}^2,{\beta}^2)$ .

to find the equation with ${\alpha}^2,{\beta}^2$ as roots put $\sqrt{x}$ in place of $x$

The equation becomes

$x-\sqrt{x}-1=0$

$x-1=\sqrt{x}$

Squaring

$x^2-3x+1=0$

now ${\alpha}^2$ satisfies this equation

$\Rightarrow {\alpha}^4-3{\alpha}^2+1=0$

Multiplying by ${\alpha}^{2n}$

${\alpha}^{2n+4}-3{\alpha}^{2n+2}+{\alpha}^{2n}=0$

Now similar as above)