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Let $f = a_nX^n+\cdots+a_1X\pm p \in \mathbb{Z}[X]$ with $\sum_{i=1}^n |a_i| < p$. Show that $f$ is irreducible in $\mathbb{Q}[X]$.

Hint: Show that every root of $f \in \mathbb{C}$ has modulus greater than $1$ and consider leading and constant terms of a factor of $f$.

I have been able to show that every roots has modulus greater than $1$. But I am not able to go any further? Please help.

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If $f(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots+a_1x+(-1)^up$ has a solution in $\mathbb{Q}$ which is $r/s$ with $gcd(r,s)=1$, then $a_nr^n+a_{n-1}r^{n-1}s+\ldots+a_1rs^{n-1}+(-1)^ups^n=0\\\Rightarrow r(a_nr^{n-1}+a_{n-2}r^{n-1}s+\ldots+a_1s^{n-1})=-(-1)^ups^n\\\Rightarrow r|p.$

If $p$ is prime then we have a contradiction.

If $p$ is not prime then we have a counterexample $f(x)=x^2+x-6$

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  • 2
    $\begingroup$ How to show that $f(x)$ has no irreducible factor? With no roots doesn't means irreducibility. $\endgroup$ – Mittal G Nov 22 '18 at 14:47

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