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Is there a degree-5 polynomial $ p\left(t\right) = a_0 + a_1 \cdot t + a_2 \cdot t^2 + a_3 \cdot t^3 + a_4 \cdot t^4 + a_5 \cdot t^5 $ with integer coefficients $ a_j $, such that $ p\left(t\right) $ is irreducible over rationals, but the second iterate $ p\left(p\left(t\right)\right) $ is not irreducible over rationals?

This is a special case of an earlier question I asked on Math.SE. According to an answer there, such polynomials are known as polynomials with newly reducible second iterate. The same answer by Sil gives this example of a degree-5 polynomial with rational coefficients: $$ \begin{align} p\left(t\right) &= t^5 + 10 t^4 + 40 t^3 + 80 t^2 + \frac{1232}{15} t + \frac{15904}{465} \\ p\left(p\left(t\right)\right) &= \left( t^5 + 10 t^4 + 40 t^3 + 80 t^2 + \frac{1202}{15} t + \frac{14974}{465} \right) \cdot Q\left(t\right) \end{align} $$ where $ Q\left(t\right) $ is shorthand for a degree-20 factor.

I searched for an example with integer coefficients using a script, and found no examples where $ \sum_j {a_j^2} < 100 $.

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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

1 Answer 1

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I also did a brute-force search with $|a_i| \leq 6$ for all $i$ and found $$ p(t) = -3t^5 - 3t^4 + 4t^3 + t^2 - 3t + 1, $$ where $p(p(t))$ has the factor $$ 3t^5 - 3t^4 - t^3 + 4t^2 - 3t + 1. $$

The only other example I found in this range was $$ p(t) = -3t^5 + 3t^4 + 4t^3 - t^2 - 3t - 1. $$

Anyways, this either means that I misunderstood the question, or there is an error in your script ;).

I will do a slightly bigger search today to maybe find a monic example.


Added by Peter Mueller (see comments below): Here is the code which does the parallelized search. Save it to a file like par.sage with filename ending in .sage and call sage par.sage. The program will ask for the number of cores to be used, and then starts that many SageMath processes in parallel, logging the progress and the findings to the single file par.log:

import os
from itertools import product
import fcntl

name = sys.argv[0]

def write(text):
    with open(name[:-4] + 'log', 'a') as f:
        fcntl.flock(f, fcntl.LOCK_EX)
        f.write(text + '\n')
        f.flush()
        os.fsync(f.fileno())
        fcntl.flock(f, fcntl.LOCK_UN)
        
if len(sys.argv) == 3:
    N = int(sys.argv[1])
    r = int(sys.argv[2])
    R.<t> = QQ[]
    m = 0
    while True:
        m += 1
        write(f'core = {r}, coefficient bound = {m}')
        l = (range(1, m+1),) + (range(-m, m+1),)*5
        for tup in product(*l):
            if sum(tup) == 0 or sum(tup) % N != r:
                continue
            if tup[-1] == 0 or (min(tup) > -m and max(tup) < m):
                continue
            P = R(tup)
            if not P(t=P).is_irreducible():
                if P.is_irreducible():
                    write(str(P))
else:
    N = int(input('Enter the number of cores to use for the search: '))
    for r in range(N):
        os.system(f'nohup sage {name} {N} {r} &')
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    $\begingroup$ I extended the search to $|a_i| \leq 9$ for all $i$, but no new examples where found. $\endgroup$ Commented Feb 14 at 23:19
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    $\begingroup$ Searching in a much larger range yielded another example: $p(t)=-4t^5 + 8t^4 + 8t^3 - 18t^2 - t + 5$ is irreducible, but $p(p(t))$ is divisible by $4t^5 - 4t^4 - 10t^3 + 8t^2 + 3t - 2$. $\endgroup$ Commented Feb 21 at 7:58
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    $\begingroup$ One more example: $p(t)=4t^5 + 12t^4 + 20t^3 + 17t^2 + 8t + 1$. Now $p(p(t))$ splits into irreducibles over $\mathbb Q$ of degrees $10$ and $15$. The degree $10$ factor is $8t^{10} + 48t^9 + 152t^8 + 310t^7 + 444t^6 + 460t^5 + 347t^4 + 188t^3 + 70t^2 + 16t + 2$. $\endgroup$ Commented Feb 21 at 17:26
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    $\begingroup$ @PeterMueller, were you able to search a larger range because you have more computing power, or did you use some tricks to speed up the search (if so, which)? $\endgroup$ Commented Feb 22 at 8:06
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    $\begingroup$ The Sage code is too long for the comment field, and it is not worth a new "answer". If you don't mind, I can append it at the end of your answer. $\endgroup$ Commented Feb 22 at 9:47

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