How can I show for some transcendental number $\alpha \in \mathbb{C}$, that two distinct polynomials of the form $b_{n}\alpha^n + \cdot \cdot \cdot + b_{2}\alpha^2 + b_{1}\alpha$ and $a_{m}\alpha^m + \cdot \cdot \cdot + a_{2}\alpha^2 + a_{1}\alpha$ can never equal each other?
"Non-equality" of distinct polynomials with the same transcendental argument
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2You should probably make explicit the assumption that the coefficients $a_i,b_i$ are algebraic numbers, since otherwise the assertion is not true. – 2011-02-21
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0@Peter: of course, you're right. But I assume the OP meant transcendental over $\mathbb Q$, which is what unadorned "transcendental" usually means. – 2011-02-22