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.

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.
