Find the value of $a^4-a^3+a^2+2$ when $a^2+2=2a$ Find the value of $a^4-a^3+a^2+2$ when $a^2+2=2a$
My Attempt, 
$$a^4-a^3+a^2+2=a^4-a^3+2a$$
$$=a(a^3-a^2+2)$$
What's next?
 A: $a^4-a^3+a^2+2=(a^2+a+1)(a^2-2a+2)=0$.
We can check also $a=1+i$ and $a=1-i$.
A: Obviously $a \not = 0$. Then we have $a^4 = 2a^3 - 2a^2$ by multiplying the condition by $a^2$, so the equation becomes: $a^3 - a^2 + 2$. Similarly $a^3 = 2a^2 - 2a$, so the equation becomes: $a^2 - 2a + 2 = 0$
A: Hint $\ $ We seek the value of $\,f(a)\,$ given that $\, g(a) = a^2-2a+2 = 0.\,$ By the Polynomial Division Algorithm we can write  $\ f = q\,g + r\,$ so $\, g=0\,\Rightarrow\, f = r = f\bmod g.\,$ So it suffices to compute the remainder $\, r = (f\bmod g).\,$ Doing so easily yields $\,f = r = 0.$
Because $\, g = 0\,$ we can perform all arithmetic modulo $g.\,$ This is the point of modular arithmetic - when we are interested only in the remainder then we can ignore the quotients. This is exactly what is done (implicitly) in Stefan4024's answer. 
A: Probably the most "uninspired" (and therefore easy to arrive at) solution is to solve the quadratic to give $a = 1 \pm i$
Then substitute one of these solutions into the quartic expression and show that it vanishes. There's no need to test the other (complex conjugate) root by substituting it into the quartic as it has real coeffients and will therefore be zero in the other case as well.
