This question takes the factorisation of a polynomial $p(x)=q(x)r(x)$, where $p$ (and for my purpose here $q$ and $r$) have integer coefficients and asks if the maximum absolute value of the coefficients of $q,r$ can ever be greater than the maximum absolute value of the coefficients of $p$.

That is answered by a factor of $x^{105}-1$ which has a coefficient of absolute value $2$ and there are other examples of higher degree amongst the cyclotomic polynomials.

Also note that $x^4+1=(x^2+1)^2-2x^2$ has the factorisation $(x^2+\sqrt 2 x + 1)(x^2-\sqrt 2 x +1)$ which isn't integral, but does suggest that lower degree examples may exist.

But what is the lowest degree of an integer polynomial where $q,r$ have integer coefficients which are not bounded by the maximum absolute value of the original polynomial?

(apologies for the clumsy explanation - I couldn't find a neater way to ask what I wanted).


I supposed that @Ewan Delanoy's example were minimal, and on trying to prove that, I found the following very similar degree-3 factorization: $$X^3-X^2-X+1 = (X^2-2X+1)(X+1)$$ which must be minimal because at least one factor must have degree at least $2$ because the leading and trailing coeffs of the factors are bounded by those of the product.

Kudos go to Ewan, as without his example, I would not have constructed the other one.

P. S.: If you want a factorization into irreducibles, consider $$X^3+2X^2-2X-1 = (X^2+3X+1)(X-1)$$


What about

$$ X^4+X^3+X+1=(X^2+2X+1)(X^2-X+1) $$

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    $\begingroup$ You could also have $x^4+x^3-x^2-1 = (x-1)(x^3+2x^2+x+1)$. If I'm understanding arxiv.org/abs/0904.3057 correctly it claims that this is the unique quartic with height 1 and an irreducible factor of height greater than 1, in which case your example is a correction. $\endgroup$ – Peter Taylor Oct 31 '14 at 17:21
  • $\begingroup$ @PeterTaylor My example is not irreducible (see ccorn’s answer for the version with common factors cleaned up) $\endgroup$ – Ewan Delanoy Oct 31 '14 at 17:53
  • $\begingroup$ Ah, that explains it. Oops. $\endgroup$ – Peter Taylor Oct 31 '14 at 17:55
  • $\begingroup$ Being new to awarding bounties I couldn't split the reward as I would have done had I been able. Thanks for this answer and the effort which went into it - it was very helpful and enlightening. $\endgroup$ – Mark Bennet Nov 5 '14 at 16:44
  • $\begingroup$ @PeterTaylor Thanks for that reference which takes the subject further. $\endgroup$ – Mark Bennet Nov 5 '14 at 16:46

No, there is no lowest degree of factorization if coeffecients can be irrational. Every binomial can be written in the form $a+b$ or $a-b$ ,both of which can be factorized.

$a+b=(\sqrt a+\sqrt b)(\sqrt a-\sqrt b)$

$a-b=(\sqrt a +\sqrt b-\sqrt (2\sqrt (ab))(\sqrt a+\sqrt b+\sqrt (2\sqrt (ab))$

Each factor found can further be factorized using the same identities. $(\sqrt a+\sqrt b+\sqrt (2\sqrt (ab))$ will have to be written as $(\sqrt a+\sqrt b)+(\sqrt (2\sqrt (ab))$ for the $a+b$ identity to be used. (and similarly for the other factor.

If factors have to have to be integers, the lowest possible degree of a factor is $1/2$, that is when the factor is of the form $a+b$, it can be further factorized using the identity exactly once, keeping the coeffecients as integers. However, there has to be atleast 1 factor with degree 1.

  • $\begingroup$ I gave an example with square roots in it, and pointed out it wasn't integral. I am specifically looking for factorisations in $\mathbb Z[x]$ (if you want it put formally). Your example is interesting, but misses the point of my question. $\endgroup$ – Mark Bennet Nov 1 '14 at 15:51
  • $\begingroup$ You have mixed up your factorisations, I think. $\endgroup$ – Mark Bennet Nov 1 '14 at 15:51
  • $\begingroup$ Sorry, I didn't get the exact question. The factorization appears to be right to me, though. $\endgroup$ – ghosts_in_the_code Nov 3 '14 at 16:26

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