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Writing the few lines below in PARI/GP, one easily checks that the polynomials


are all irreducible over $\mathbb Z$ when $a_0,a_1$ and $a_2$ are integers between $0$ and $20$. Is there a computer-free (or more illuminating) proof of this fact ?



for(a2=0,20,for(a1=0,20,for(a0=0,20, vv1[(21^2)*a2+21*a1+a0+1]=ff3(a2,a1,a0);););)


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irreducibility of some of these polynomials you can prove with Eisenstein's criterion –  pedja Nov 12 '11 at 15:11
@ pedja : right, it works for some but not all. I'm looking for a criterion like Eisenstein's that might apply to all of them at once. –  Ewan Delanoy Nov 12 '11 at 16:10
I know for Perron's and Cohn's criterion but I don't think that you can apply them on these polynomials... –  pedja Nov 12 '11 at 16:17

2 Answers 2

up vote 4 down vote accepted

Sketch. Because the constant term is so large relative to the other terms, possible integer roots of such a polynomial lie in the relatively small interval $[160, 165]$ far to the right of the origin. The fact that $X_0$ is a root is equivalent to $4019680$ having a representation in base $X_0$ with particularly large coefficients, and by writing $4019680$ in the corresponding bases we see that it doesn't have such representations.

Proof. Reducibility is equivalent to the existence of an integer root. By inspection the polynomial can have at most one real root $X_0$, somewhere past $\sqrt[3]{4019680} = 159.0000131...$. By setting $a_0 = a_1 = a_2 = 20$ we check numerically that $X_0 < 166$, so the only possible roots are $X_0 = 160, 161, 162, 163, 164, 165$. Any integer root must divide the constant term, but: $$4019680 \equiv 0 \bmod 160$$ $$4019680 \equiv 154 \bmod 161$$ $$4019680 \equiv 136 \bmod 162$$ $$4019680 \equiv 100 \bmod 163$$ $$4019680 \equiv 40 \bmod 164$$ $$4019680 \equiv 115 \bmod 165.$$

Since $0 \le a_0 \le 20$, it is impossible for $X_0$ to divide the constant term unless $X_0 = 160, a_0 = 0$ or $X_0 = 161, a_0 = 7$. In either case, after dividing out by $X_0$, $X_0$ must divide the new constant terms, which are $-(25123+a_1), -(24967+a_1)$ respectively. But $$25123 \equiv 3 \bmod 160$$ $$24967 \equiv 12 \bmod 161$$

and again since $0 \le a_1 \le 20$, this is impossible.

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Very nice Qiaochu! I have the bad habit of relying too much on my computer, so I tend to forget simple proof techniques like this one. –  Ewan Delanoy Nov 12 '11 at 18:24

If you are really interested only in $0 \leq a_i \leq 20$, write the polynom as product of two polynoms of degree 1 and 2 with unknown coefficients:

$(A x^2 + B x + C)(Dx + E) = ADx^3 + (BD+AE)x^2+(BE+CD)x+CE$

If I did it correct, this yields for any B, C, E:

  • $E \cdot C = a_0 - 4019680$
  • $C + B\cdot E = a_1$
  • $E+B = a_2$

But these euqations are not solveable for $0 \leq a_i \leq 20$.

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I understand nothing in that solution. What are $B,C$ and $E$ ? Also, a polynom of degree 3 is irreducible iff it has no root. –  Ewan Delanoy Nov 12 '11 at 16:09
Sorry, I improved the answere a little bit. –  lumbric Nov 12 '11 at 16:24

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