Prove that a polynomial is irreducible over $\mathbb Z$

Question

How do you prove that a polynomial is irreducible, and in this case that $(x - a)(x - b) - 1$ is irreducible over $\mathbb Z$?

EDIT: $a≠b$

Answer 1

Note that a integer polynomial is irreducible over $\mathbb Z$ iff it is over $\mathbb Q$ and a quadratic polynomial is irreducible over $\mathbb Q$ iff it has no zeros. The zeros of your polynomial over $\mathbb C$ are $$ x_{1/2} = \frac{a+b}2 \pm \sqrt{\frac{(a+b)^2}4 - ab + 1} = \frac{a+b}2 \pm \sqrt{\frac {(a-b)^2 + 4}4} $$ $x_{1/2}$ are rational if $(a-b)^2 + 4$ is a square in $\mathbb N$, but $(a-b)^2$ is an integer square, and two squares do not have difference $4$. That's not right (see the comments, thanks Arthur and arbautjc), but there is only one case, namely $(a-b)^2 = 0$. If $a\ne b$, your polynomial is irreducible, in case $a=b$, we have $$ (x-a)(x-a)^2 - 1 = \bigl(x - (a+1)\bigr)\bigl(x-(a-1)\bigr) $$

Answer 2

Hint:

Suppose this has integer roots then, we would have :

$(x-a)(x-b)=1\Rightarrow (x-a)=(x-b)=1 \text { or } (x-a)=(x-b)=-1$

Is this possible?

Caution : $a\neq b$