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Claim : Let $0<a<b<c<d$ be integers. Assume that $d$ is prime. Then, the polnomyial $$f(x)=ax^3+bx^2+cx+d$$ is irreducible over $\mathbb Q$

How can I prove this result without using $f(x)=(rx^2+sx+t)\cdot (ux+v)$ and comparing the coefficients, which requires several casea and is not quite easy ?

I am looking for a compact proof using finite fields or criterions like Eisenstein or similar approaches.

Since the degree is $3$, it is sufficient to prove that the polynomial has no rational root. So, maybe the rational root theorem can help, but I did not find a proof using this approach which is easier than comparing coefficients.

Does anyone have an idea ?

Peter
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  • Well if $p/q$ is a rational root, there are divisibility relations between $p,q, d$ and $a$. The fact that $d$ is prime and $a<d$ may contradict that. – Maxime Ramzi Mar 30 '17 at 20:35
  • @Max Try to figure it out and write an answer if you are successful. – Peter Mar 30 '17 at 20:38
  • Here : https://arxiv.org/pdf/1612.01712.pdf is a generalized criterion covering my problem. My only question : What does the part "$a_{k-1}<a_k<a_{k+1}$ in connection with "$0\le k\le n-1$" mean ? Does it mean that $a_0<a_1$ is sufficient, whereas $a_1<a_2$ is not ? Or does it mean that we have at least one "<"-sign in the inequality-chain ? – Peter Apr 01 '17 at 23:05

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