Finding the value of $ax^4+by^4$ 
If $\quad a+b=23 , \quad ax+by=79,\quad ax^2+by^2=217,\quad
 ax^3+by^3=691\quad$ find the value of $ax^4+by^4$.

Here is my attempt:
$$(a+b)(x+y)=ax+by+ay+bx\rightarrow 23(x+y)=79+(ay+bx)$$
$$(ax+by)(x+y)=ax^2+by^2+axy+bxy\rightarrow79(x+y)=217+23 xy$$
In each equation I have two unknowns it seems that doesn't work.
 A: Your method is not in vain. We have:
$$(ax+by)(x+y) = ax^2 + by^2 +axy + bxy \to 79(x+y) = 217 + 23xy$$
$$(ax^2 + by^2)(x+y) = ax^3 + by^3 + ax^2y + bxy^2 \to 217(x+y) = 691 + 79xy$$
hence $x + y = 1, xy = -6$. Now:
$$(ax^3 + by^3)(x+y) = ax^4 + by^4 + ax^3y + bxy^3 = ax^4 + by^4 + xy(ax^2 + by^2)$$
$$691\times 1 = ax^4 + by^4 -6 \times 217$$
thus $ax^4 + by^4 = 1993$.
A: Hint:
$$(a+b)(ax^2+by^2)-(ax+by)^2=\cdots=ab(x-y)^2\ \ \ \ (1)$$
$$(a+b)(ax^3+by^3)-(ax+by)(ax^2+by^2)=\cdots=ab(x-y)^2(x+y)\ \ \ \ (2)$$
On division, we get $x+y$
$$(a+b)(ax^3+by^3)-(ax^2+by^2)^2=\cdots=abxy(x-y)^2\ \ \ \ (3)$$
$(1)/(3)$ will give $xy$
$$(a+b)(ax^4+by^4)-(ax+by)(ax^3+by^3)=\cdots=ab(x-y)^2((x+y)^2-xy)\ \ \ \ (4)$$
$(4)/(3)$ will give $$\dfrac{23(ax^4+by^4)-79\cdot691}{23(691)-217^2}=\dfrac{xy}{(x+y)^2-xy}$$
Now replace the values of $x+y$ and $xy$
A: At the core of contest problems like

Given $x^n+1/x^n=p$, evaluate $x^m+1/x^m$


Given $x^n+y^n=p$, evaluate $x^m+y^m$

for most likely, integers $m,n$, lies the recurrence relation
$$x^{n+1}+y^{n+1}=(x+y)(x^n+y^n)-xy(x^{n-1}+y^{n-1})$$
Our question seems a bit generalized but in the same spirit, the following recurrence relation can be formed :
$$ax^{n+1}+by^{n+1}=(x+y)(ax^{n}+by^{n})-xy(ax^{n-1}+by^{n-1})$$
Thus the problem reduces to finding the linear coefficients $(x,y)$ or $(x+y,xy)$.
To ease the calculation, we do the following :
$$217=79(x+y)-23xy \tag{1}$$
$$691=217(x+y)-79xy \tag{2}$$
Lets do $3\times(1)-(2)$. We get
$$-40=20(x+y)+10xy$$
$$\Rightarrow (x+2)(y+2)=0$$
Thus one of $x,y$ is $-2$ but not both. (Why?) Putting $x=-2$ in $(1)$ gives $y=3$. Our recurrence relation is
$$ax^{n+1}+by^{n+1}=(ax^{n}+by^{n})+6(ax^{n-1}+by^{n-1})$$
which means this ladder goes up in following manner :
$$79+6\cdot 23=271$$
$$271+6\cdot 79=691$$
$$691+6\cdot 217=1993$$
Thus we have observed a systematic method to tackle these type of problems.
A: Answer to OP's comment:
For solving the following system:

$$79(x+y)=23xy+217$$$$ 217(x+y)=79xy+691$$

We first eliminate $(x+y) $ from both equations:
$$ (x+y) = \frac{1}{79} (23 xy + 217)\tag{1}$$
And,
$$ x+y = \frac{1}{217} (79 xy + 691) \tag{2}$$
We get:
$$  \frac{1}{217} (79 xy + 691)= \frac{1}{79} (23 xy + 217)$$
We get the result
$$ y = \frac{-6}{x} \tag{3}$$
Now we can substitute (3) into (1), and then if we multiply the whole equation by $x$ , we can solve for the quadratic in x
