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# non costant

+1
198
5
+874

Find the nonconstant polynomial $$P(x)$$ such that

$$P(P(x)) = (x^2 + x + 1) P(x)$$

Dec 28, 2018

#1
+102441
0

I do not know but it looks interesting. I will be interested in the answer when it comes

Dec 29, 2018
#2
+5225
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it's pretty clear P(x) can't be an odd degree so we'll try a 2nd degree polynomial

$$\text{Letting }P(x) = c_0 + c_1 x + c_2 x^2\\ \text{then grinding out the expansions of each side and equating like power of x coefficients}\\ \text{we get two solutions}\\ P(x) = 0\\ P(x) = x+x^2$$

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Dec 29, 2018
edited by Rom  Dec 29, 2018
#3
+102441
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Thanks Rom,

I will have to think about that one :)

Melody  Dec 29, 2018
#4
+5225
+3

Maybe you can follow this.

Rom  Dec 29, 2018
#5
+102441
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Thanks Rom.

I will look :)

Melody  Dec 29, 2018