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Hello,

Let $p(x,y) = \sum_{m=1}^M\sum_{n=1}^N a_{m,n}x^{m-1}y^{n-1}$ be a bivariate polynomial where $\{a_{m,n}\}$ are complex.

If $(x_k, y_k), k=1,2,\cdots, MN-1$ are roots of $p(x,y)=0$ where $|x_k|=|y_k|=1$ for $k=1,2,\cdots,MN-1$.

How can I show that there exists another different solution, that is, $p(x_{MN}, y_{MN})=0$ where $|x_{MN}|=|y_{MN}|=1$ and $(x_{MN}, y_{MN}) \neq (x_{k}, y_{k})$, $k=1,2,\cdots,MN-1$?

Thanks a lot.

Substituting $x = \frac{1-t^2}{1+t^2} + i\frac{2t}{1+t^2}$ and $y = \frac{1-s^2}{1+s^2} + i\frac{2s}{1+s^2}$ into $p(x,y)$, we get $p(t,s)$. Then $p(x,y) = 0$ implies $p_r(t,s)\triangleq\Re(p(t,s))=0$ and $p_i(t,s)\triangleq\Im(p(t,s))=0$. Then, the problem becomes:

1)What is the number of common roots of $p_r(t,s)$ and $p_i(t,s)$.

2)If there are $MN-1$ common roots, will there be another common root?

Any comments?

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  • $\begingroup$ Do you really mean for your indices to start at $1$? so all your polynomials are divisible by $xy$? $\endgroup$
    – Gerry Myerson
    Commented Jan 31, 2013 at 4:37
  • $\begingroup$ This has a certain homework flavor. Voting to close. $\endgroup$
    – Igor Rivin
    Commented Jan 31, 2013 at 5:03
  • $\begingroup$ Sorry, there is a constant in p(x,y). $\endgroup$
    – Frank
    Commented Jan 31, 2013 at 5:45
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    $\begingroup$ This is NOT a homework. I just tested the statement with numerical examples and would like to know if it is true theoretically. Can anyone offer some hints? $\endgroup$
    – Frank
    Commented Jan 31, 2013 at 6:43
  • $\begingroup$ Are there any properties of unimodular roots ($|z|=1$) of a bivariate polynomial equation? For example, the number of unimodular roots, the space containing the unimodular roots and the common unomodular roots of two bivariate polynomial equations? Thanks. $\endgroup$
    – Frank
    Commented Jan 31, 2013 at 7:46

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