Find all polynomials $g(x)$ with real coefficients with the property $$g(x)g(x-1)=g(x^2).$$
My try: I found $$g(x)=(x^2+x+1)^n$$ satisfies the condition; maybe there are other solution? If so, how to prove it (and/or find them)?
Thank you.
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2Hint: What can you say about the roots of the polynomial? – Calvin Lin Oct 18 '13 at 14:34
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$g=0$ and $g=1$ are solutions. – lhf Oct 18 '13 at 15:02
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Hint: What can you say about the roots of the polynomial?
(The following are steps. Fill in the rest of the details yourself. It should be clear why this works. I removed as much text as possible due to OP's request in comments.)
If $g(x) = g$ is constant, then $g^2 = g$.
Otherwise, $\deg g \geq 1$. Let $g(\alpha) = 0 $.
Step 1: $g( \alpha) = 0 \Rightarrow | \alpha | = 0 $ or 1.
Step 2: If $\alpha = 0 \Rightarrow g(n^2 ) = 0\, \forall n \in \mathbb{N}$. $\Rightarrow \Leftarrow$
Step 3: If $|\alpha |= 1 \Rightarrow \exists n \in \mathbb{N}, \alpha^n = 1$.
Step 4: If $\alpha^n = 1$, then for $\beta = (\alpha + 1)^2$, we have $g(\beta) = 0 $.
Step 5: $\alpha = ??$
Step 6: If $(x-\alpha)^m \mid g(x) \Rightarrow (x-\beta)^m \mid g(x).$
Step 7: Hence $g(x) = (x^2 + x+1)^n$.
Calvin Lin
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Hello,My English is not good,plese don't say more text ,can you take more mathematical,That I can understand – math110 Oct 18 '13 at 14:45
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@chinamath Sure, they are all in math statements now. Fill in the details yourself. – Calvin Lin Oct 18 '13 at 14:50
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Thank you,can you help me see this question: http://math.stackexchange.com/questions/530887/how-find-this-maxium-s-delta-abc?rq=1 – math110 Oct 18 '13 at 15:18