Given an integral polynomial of degree strictly greater than one, completely decompose it into a composition of integral polynomials of degree strictly greater than one.


  • An integral polynomial is a polynomial with only integers as coefficients.
  • Given two polynomials p and q the composition is defined by (p∘q)(x):=p(q(x)).
  • The decomposition of an integral polynomial p is a finite ordered sequence of integral polynomials q1,q2,...,qn where deg qi > 1 for all 1 ≤ i ≤ n and p(x) = q1(q2(...qn(x)...)), and all qi are not further decomposable. The decomposition is not necessarily unique.
  • You can use e.g. lists of coefficients or built in polynomial types as input and output.
  • Note that many builtins for this task actually decompose the polynomials over a given field and not necessarily integers, while this challenge requires a decomposition integer polynomials. (Some integer polynomials might admit decomposition into integer polynomials as well as decomposition that contain rational polynomials.)


x^2 + 1
[x^2 + 1] (all polynomials of degree 2 or less are not decomposable)
x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6 x - 1
[x^3 - 2, x^2 - 2x + 1]
x^4 - 8x^3 + 18x^2 - 8x + 2 
[x^2 + 1, x^2 - 4x + 1]
x^6 + x^2 + 1
[x^3 + x + 1, x^2]
[x^2, x^3]
x^8 + 4x^6 + 6x^4 + 4x^2 + 4 = (x^2 + 1)^4 + 3
[x^2 + 3, x^2, x^2 + 1]
x^6 + 6x^4 + x^3 + 9x^2 + 3x - 5
[x^2 + x - 5, x^3 + 3*x], [x^2 + 5*x + 1, x^3 + 3*x - 2]

Use Maxima for generating examples: Try it online!

Some decomposition algorithms can be found here and here.


2 Answers 2


Pari/GP, 84 bytes


Based on the algorithm described here.

Try it online!

  • 1
    \$\begingroup\$ Do you check (or filter out) whether you actually get a decomposition into integral polynomials? (I'm asking because the algorithms in the linked paper describe the factorization over some field, and I don't know any Pari/GP.) \$\endgroup\$
    – flawr
    Mar 12, 2018 at 12:10
  • 1
    \$\begingroup\$ @flawr I'm using the second algorithm in the paper, which always returns integral polynomials when the input is integral. In fact, the divisors function in Pari/GP always returns primitive polynomials when it takes an integral polynomial. It can be proved that if p=q∘r, where p and r are integral, and r is primitive with r(0)=0, then q must also be integral. Here p, q, r correspond to f, g, h in the paper. \$\endgroup\$
    – alephalpha
    Mar 12, 2018 at 14:26

Wolfram Language (Mathematica), 29 bytes


Try it online!

I have the example set up here to compose a random polynomial from random quadratics (or less), expand it out, and then try to decompose it.

It's necessary to complicate the polynomial with dummy variable (a) since the built-in will not attempt to decompose a monomial.

I notice that the answer often has much larger coefficients than in the original composition, but they are indeed always integers.

  • \$\begingroup\$ Where did you find the information that Decompose[] will always return integral polynomials (if fed with integer polynomials)? When discussing in chat recently we could not find anything about that. \$\endgroup\$
    – flawr
    Mar 12, 2018 at 16:09
  • 1
    \$\begingroup\$ Do Options@Decompose and it will tell you {Modulus->0}. Now look up Modulus and you'll see "The setting Modulus->0 specifies the full ring [DoubleStruckCapitalZ] of integers. " \$\endgroup\$ Mar 12, 2018 at 16:24
  • \$\begingroup\$ Ah that is nice, thanks for elaborating! \$\endgroup\$
    – flawr
    Mar 12, 2018 at 16:47

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