http://en.wikipedia.org/wiki/AKS_primality_test
How can I interpret what the "mod n" means?
I have watched the Numberphile video on the AKS primality test, and based on that, I am assuming that "mod n" means the remainder when dividing by each of the coefficients.
$$ (x - 1)^5 \equiv (x^5 - 1) \pmod 5 $$ $$ x^5 - 5x^4 + 10x^3 - 10x^2 + 5x - 1 \equiv x^5 - 1 \pmod 5 $$ $$ 1x^5 - 0x^4 + 0x^3 - 0x^2 + 0x - 1 \equiv 1x^5 - 1 $$ $$ x^5 - 1 \equiv x^5 - 1 $$
Is my assumption correct?
On
Your assumption is correct if $x$ (and $a$?) is an indeterminate.
To add a bit of complication, if $x$ is a variable that can take integer values, additional congruences are possible, e.g., $x^2+x\equiv 0\pmod 2$. But nevertheless, any result obtained by reducing coefficients modulo the module remains corrct also for the second interpertation.
This is usually covered in more detail in an introductory number theory or abstract algebra textbook, and is referred to as "modular arithmetic."
Writing that two quantities are congruent/equivalent$\pmod n$ (this is said "modulo $n$") means that the two quantities produce the same remainder when divided by $n$. Thus, we can write $15\equiv -1\equiv 7\pmod 8$.
We often want to work with polynomials, though; in this case, we can simplify expressions (as you have in the question) by reducing each of the coefficients of the polynomial to their "residues" (remainders) modulo $n$.
The big "watchout" in modular arithmetic is that, although we can multiply, add, and subtract freely, division isn't always guaranteed. Instead, we work with "inverses" of a number. Long story short, an integer $a$ is the inverse of $x$ modulo $n$ if (and only if) $ax\equiv 1\pmod{n}$. We write $a\equiv x^{-1}\pmod n$. It can be shown that the inverse always exists if $x$ and $n$ are coprime.
All that to say: yes, your approach in the question is correct.