<s>[INST] Does there exists a polynomial $Q(x,y)$ such that $x-1=Q(x^2-1,x^3-1)$. I have a question: Does there exists a polynomial $Q(x,y)$ such that $x-1=Q(x^2-1,x^3-1)$.
I did as following: Let $Q(x,y)=\sum\limits_{k=0}^n\sum\limits_{i+j=k}a_{ij}x^iy^j$. Then I could find some coefficients: the coefficient of $(x^2-1)$ is $0$,coefficient of $x^3-1$ is $-\frac13$, ...
Now I get stuck. Is the proposition true? How to prove it?
 [/INST] The polynomials $x^2-1,x^3-1,(x^2-1)^2,(x^2-1)(x^3-1),(x^2-1)^3,(x^3-1)^2,(x^2-1)^2(x^3-1),(x^2-1)^4,(x^2-1)(x^3-1)^2$ are $9$ in number and all of degree at most $8$. I suspect they span the $8$-dimensional space of polynomials of degree at most $8$ vanishing at $x=1$. If that's right, then $x-1$ will be a linear combination of them. Worth a try, I think. 
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