Question: In $\mathbb{F}_3[x]$ write, if possible, the polynomial $1$ in the form: $$f(x)p(x)+g(x)q(x)=1$$ Where $p(x)=x^3+x^2+x+2$ and $q(x)=x^3+2x^2+2$.
This is a question from Concrete Introduction to Higher Algebra (Lay) assigned in my Classical Algebra class.
I first jumped to using the EEA matrix, but ran into issues.
I then used Euclid's algorithm/long division to find that $\gcd(p(x),q(x))=-x^2+x \neq 1$ so am I correct in saying that there is no way to write the polynomial $1$ in the form stated above?
My other question was, does the fact that these polynomials are in $\mathbb{F}_3[x]$ make any tangible difference in solving this problem?
Euclidean algorithm: $$ \begin{array}{rcrll} x^3+x^2+x+2 &=& 1 &\cdot& (x^3+2x^2+2) &{}+ (2x^2+x) \\ x^3+2x^2+2 &=& 2x &\cdot& (2x^2+x) &{}+ 2 \\ 2x^2+x &=& (x^2+2x) &\cdot& 2 & \end{array} $$ This shows that $2 = -1$ is the GCD, but it's only defined up to a unit, so it's equivalent to the fact that $1 = 2 \cdot 2$ is the GCD, i.e. the polynomials are relatively prime.
Now, the backwards phase of the extended Euclidean algorithm will express $2$ as a $\mathbb{F}_3[x]$-linear combination of $p(x)$ and $q(x)$, which we can negate to get the Bézout identity. Begin with the penultimate line, and substitute back up the chain of equations (using $2 = -1$ freely): $$ \begin{array}{rcrcl} 2 &=& (x^3+2x^2+2) &{}+{}& x \cdot (2x^2+x) \\ &=& (x^3+2x^2+2) &{}+{}& x \cdot \bigl( (x^3+x^2+x+2) + 2(x^3+2x^2+2) \bigr) \\ &=& x \cdot (x^3+x^2+x+2) &{}+{}& (2x+1) \cdot (x^3+2x^2+2). \end{array} $$ So, $$ x \cdot p(x) + (2x+1) \cdot q(x) = 2, $$ which we can multiply by $2$ to get $$ 2x \cdot p(x) + (x+2) \cdot q(x) = 1. $$