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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?

Jie Fan
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2 Answers2

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Let $R(y,z) := Q(y-1, z-1)$. If $Q(x^2 - 1, x^3 - 1) = x-1$, then $R(x^2, x^3) = x-1$.

It's easy to see, however, that there does not exist a polynomial $R(y,z)$ satisfying $ R(x^2, x^3) = x-1$.

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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.

Gerry Myerson
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  • Thank you but it's wrong. We can observe that $(x^2-1)^3$, $(x^3-1)^2$, $x^2-1$ and $x^3-1$ are linearly dependent. – Jie Fan Mar 06 '13 at 03:47
  • How can they be linearly dependent? The first is the only one with an $x^4$ term, so its coefficient would have to be zero. That leaves the second as the only one with $x^6$, so its coefficient would be zero. That just leaves the last two, which are clearly independent. – Gerry Myerson Mar 06 '13 at 04:42
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    I think he meant to include $(x^2-1)^2$ in his list. Setting $y = x^2 - 1$ and $z = x^3 - 1$, we have a relation $(y+1)^3 = (z+1)^2$ which provides a linear relation between $y,y^2,y^3,z,z^2$. It was actually checking the two ways to reduce $x^6$ that made me discover the approach I used in my answer. –  Mar 06 '13 at 09:35
  • Yes I forgot $(x^2-1)^2$. Thanks – Jie Fan Mar 07 '13 at 07:06