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    $\begingroup$ Do we understand how this works for quadratics? cubics? $\endgroup$ Commented Jan 25 at 5:30
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    $\begingroup$ @GerryMyerson Take $$p_2(x)=4x^2-8x+5$$ implying $$p_2(p_2(x))=(8x^2-20x+13)\,(8x^2-12x+5)$$ and $$p_1(x)=x^3-4x^2+3x+1$$ implying $$p_3(p_3(x))=(x^3-5x^2+6x-1)\,(x^6-7x^5+16x^4-14x^3+6x^2-1)$$ giving you quadratic and cubic examples (found by brute force using a code written by ChatGPT). $\endgroup$ Commented Jan 25 at 12:49
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    $\begingroup$ A curious thing: looking at the list of examples on the MathSE link, I found that for all those of degree 2 $P(P)$ reducible $\implies P^*(P^*)$ reducible (actually easy to prove in general), but this is false for the examples in degrees 4 and 6, as well as for the rational example above of degree 5. ($P^*$ by the way is the RECIPROCAL, or reflected, polynomial.) Degree 3 is where it gets interesting: $P^*(P^*)$ is reducible in cases 1,4,5,6 of the 7 examples, but not in case 2,3,7. I wonder if there is a cool property of cubic polynomials that would separate the 2 sets of examples. $\endgroup$ Commented Jan 25 at 13:37
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    $\begingroup$ Ah, I just noticed! The 4 degree-3 examples I noted above, all belong to the 1-parameter family $P_a(t) = at^3+(2a-1)t^2-(2a+1)t-a$ and it's a somewhat palindromic family in that $P_a^*= P_{-a}$. $\endgroup$ Commented Jan 25 at 18:06
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    $\begingroup$ @YaakovBaruch First degree-3 examples $\left(a_0,a_1,a_2,a_3\right)$ I found (including those differing only in signs), ordered by the sum of squared coefficients: (1, 1, 0, -1), (-1, 1, 0, -1), (-1, 1, 3, -1), (1, 1, -3, -1), (-1, -3, 1, 1), (1, -3, -1, 1), (-1, 3, 4, 1), (1, 3, -4, 1), (-2, 3, 5, -2), (2, 3, -5, -2), (-2, -5, 3, 2), (2, -5, -3, 2), (-3, 5, 7, -3), (3, 5, -7, -3), (-3, -7, 5, 3), (3, -7, -5, 3), (-1, -5, 6, 8), (1, -5, -6, 8), (-10, -2, 6, 2), (10, -2, -6, 2), (-4, 7, 9, -4), (4, 7, -9, -4), (-4, -9, 7, 4), (4, -9, -7, 4), (-1, 3, 10, -8), (1, 3, -10, -8), ... $\endgroup$ Commented Jan 25 at 23:13