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# Polynomials

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Find the polynomial p(x) such that p(p(x)) = xp(x) + x^2 + 3x + 1.

Apr 20, 2022

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Let $$d = \deg (p) \in \mathbb Z^+$$.

Note that $$\deg(p\circ p) = d^2$$ and $$\deg(x\cdot p + x^2 + 3x + 1) = \max(d + 1, 2)$$

Now we eliminate the case when p(x) is constant to ensure that $$\max(d + 1, 2) = d + 1$$.

(Because if p(x) = c, c = cx + x^2 + 3x + 1 for any x, which is impossible.)

Comparing the degrees on both sides gives $$d^2 = d + 1$$, which has no solution in $$\mathbb Z^+$$.

Therefore, there are no such polynomial p(x) with p(p(x)) = xp(x) + x^2 + 3x + 1.

Apr 20, 2022