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Suppose $f(x):=a_0+a_1x+\cdots+a_nx^n$ is a polynomial in $\mathbb{Z}[x]$ and $|a_i|\leq M$ for each $i=0,\ldots ,n$. Now suppose $g(x)$ is a factor of $f(x)$ in $\mathbb{Z}[x]$, then is it possible to get a bound on the coefficients of $g(x)$ in terms of $M$ i.e. if $g(x)=\sum_{i=0}^mb_ix^i$ then does there exist some $M^\prime $, which depends only on $M$, such that $|b_i|\leq M^\prime$ for all $i=0,\ldots ,m$ ?

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    Are $n$ and $m$ fixed, or you allow dependence on these parameters?2012-10-03
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    It is quite clear that M' =< M.2012-10-03
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    Yes $M^\prime$ may depend on $n$ and $m$2012-10-03
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    If you take x = 1 and factor , what can you conclude ?2012-10-03
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    @mick: taking $x=1$, absolute value of the sum of the coefficients of $g$ is less than the absolute value of the sum of the coefficients of $f$ which is less than $nM$, then ?2012-10-03
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    @ pritam : Assume all coefficients to be positive. if you factor f(x) into g(x)h(x) and x=1 then what can you say about the sum of the coefficients of the smallest of g(x) and h(x) ? Suppose f(1) = 100 and f(1) = g(1)h(1). If h(1) < g(1) you can conclude something.2012-10-03
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    @mick: I am not getting you, can you give me the complete answer and why should I assume that all the coefficients are positive ?2012-10-03
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    h(1) is at most 10. As for negative coefficients that makes the problem somewhat harder.2012-10-03
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    Simul-posted, without notice, to MathOverflow, http://mathoverflow.net/questions/108726/getting-a-bound-on-the-coefficients-of-the-factor-polynomial --- don't do that.2012-10-04

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