# conjecture regarding the height of polynomial's square-free part

About some time I am struggling with the following interesting problem:

There is a well-known theorem of Mignotte which says that for a polynomial $f\in\mathbb{Z}[x]$ of degree $n$ and height (coefficient size) $2^\tau$, the height of its divisors is bounded by:

$2^n\mathcal{M}(f)=\mathcal{O}(2^{n+\tau})$,

see, e.g., http://arxiv.org/abs/0904.3057

In other words, the height of polynomial's divisors can be larger than the height of a polynomial itself. Example could be:

$x^5+3x^4+2x^3-2x^2-3x-1=(x^4+4x^3+6x^2+4x+1)(x-1)$.

My conjecture is that this cannot happen for polynomial's square-free part. To be precise, for a polynomial $f\in\mathbb{Z}[x]$ of height $2^\tau$, its square-free part $f^*=f / \gcd(f,f')$ cannot have height larger than $2^\tau$.

Simple example: $f=(x-1)^3=x^3-3x^2+3x-1$ and $f^*=x-1$.

I appreciate if someone has ideas how to prove that or find a counterexample.

Why it's an interesting problem is because theoretical complexity of many algorithms from algebraic geometry use such bounds. On a similar note, modular approaches use these bounds to estimate the number of primes needed for computation.

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The squarefree part of $$(x^4+7x^3+8x^2+7x+1)(x-1)^2=x^6+5x^5-5x^4-2x^3-5x^2+5x+1$$ is $$(x^4+7x^3+8x^2+7x+1)(x-1)=x^5+6x^4+x^3-x^2-6x-1$$ which has greater height. It's clear the coefficients and the degree can be varied greatly to get as many counterexamples as desired.