<s>[INST] Finding the inverse of $x^3+x^2+1$ in $\Bbb F_2[x]/(x^4+x^2)$ with the Euclidean algorithm My first thought was successful: $x^4+x^2=x^2(x^2+1)$ and $x^3+x^2+1=x^2(x+1)+1$ so it is its own inverse because $(x^2(x+1)+1)^2\equiv x^4(x+1)^2+1\equiv x^4(x^2+1)+1\equiv1.$
The given solution claims to use the Euclidean algorithm, succintly, but I don't quite get it:

$x^4+x^2=(x^3+x^2+1)(x+1)+(x+1); \\ x^3+x^2+1=x^2(x+1)+1$
hence $1=(x^3+x^2+1)(x^3+x^2+1)+x^2(x^4+x^2)$

What is the logic in the above lines? How does it work in general?
 [/INST] For integers, if you're trying to find an inverse of $a$ mod $n$ using Euclidean algorithm, you would divide $n$ by $a$, find the remainder $r_1$, then divide $a$ by $r_1$, find the remainder $r_2$; divide $r_1$ by $r_2$, etc., until the remainder becomes 1. Here you're looking for the inverse of $x^3+x^2+1$ mod $x^4+x^2$, so you divide $x^4+x^2$ by $x^3+x^2+1$, find the remainder $x+1$, then divide $x^3+x^2+1$ by this remainder $x+1$, this time the remainder is already 1.
To find the inverse, you work backwards. The second equation tells you 
$$1 = (x^3+x^2+1) - (x^2(x+1))$$
By the first equation, 
$$x+1 = [x^4+x^2] - [(x^3+x^2+1)(x+1)]$$
Plug this into the above equation:
$$1=(x^3+x^2+1)- x^2(x^4+x^2-(x^3+x^2+1)(x+1)) \equiv (x^3+x^2+1)(1+x^3+x^2)$$
In short, you're doing exactly the same thing as when you solved congruence relations of integers using Euclidean algotithm.
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