<s>[INST] Algebraic shortcut to find $a^n + b ^n$? Recently, I found this problem online:

Given $a+b=1$ and $a^2+b^2=2$, find $a^7+b^7$.

Although I could've solved it by substituting the first equation into the second and then using the quadratic formula; the way the question was set up, I suspected that there was a shortcut. However, I couldn't find the solution to the problem on that website, so I'm asking here:

If you are given $a+b$ and $a^2+b^2$, is there a shortcut to finding $a^n+b^n$?

 [/INST] Note that 
$$a^{n+1}+b^{n+1}=(a^n+b^n)(a+b)-ab(a^{n-1}+b^{n-1}).\tag{1}$$
Since $(a+b)^2=1$, and $a^2+b^2=2$, we have $2ab=-1$ and therefore $ab=-\frac{1}{2}$.
It follows from (1) that
$$a^{n+1}+b^{n+1}=(a^n+b^n)(1)+\frac{1}{2}(a^{n-1}+b^{n-1}).$$
Let $f(k)=a^k+b^k$. We have obtained the recurrence
$$f(n+1)=f(n)+\frac{1}{2}f(n-1).$$
Using this recurrence, we can compute our way to $f(7)$ fairly quickly.
Remarks:  $1.$ The same strategy can be used for any given $a+b$ and $a^2+b^2$.
$2.$ To go a little faster, we could use $a^{n+2}+b^{n+2}=(a^n+b^n)(a^2+b^2)-a^2b^2(a^{n-2}+b^{n-2})$. 
$3.$ If we are in a real hurry, and $n$ is not small, there are various tricks to speed up computation. For example
$$a^{2k}+b^{2k}=(a^k+b^k)^2-2(ab)^k.$$
So we can take "giant" steps. 
</s>