$x^3+ (5m+1)x+ 5n+1$ is irreducible over $\Bbb Z$

Question

How to prove that the polynomial: $x^3+ (5m+1)x+ 5n+1$ is irreducible over the set of integers for any integers $m$ and $n$? I was trying to put $x= y+p$ for some integer $p$ so that I could apply Eisenstein's criterion. But it is very tedious to find some suitable $p$.

Answer 1

If it is not, then it can be written as a product $(x^2 + ax + b)(x + c)$, where $a,b,c$ are integers.

Expanding: $x^3 + (5m + 1)x + 5n + 1 = x^3 + (a + c)x^2 + (ac + b)x + bc$.

Then $a = -c$ and this becomes: $x^3 + (5m+1)x + 5n+1 = x^3 + (b-c^2)x + bc$

Then $c^2 = b - (5m + 1)$, so $c^3 = bc - (5m + 1)c = 5n+1-(5m+1)c \equiv 1 - c \pmod 5 $

Then $c^3 + c \equiv 1 \pmod 5 $

But this can't happen:

$0^3 + 0 = 0 \equiv 0 \pmod 5 $

$1^3 + 1 = 2 \equiv 2 \pmod 5 $

$2^3 + 2 = 10 \equiv 0 \pmod 5 $

$3^3 + 3 = 30 \equiv 0 \pmod 5 $

$4^3 + 4 = 68 \equiv 3 \pmod 5 $

Answer 2

You can show that it is irreducible modulo 5 (over $\mathbb{Z}/5$) and therefore irreducible over $\mathbb{Z}$. A degree three polynomial is irreducible iff it has no roots. Check that it has no roots in $\mathbb{Z}/5$.