Proving a general reasult in polynomials. If $f(x) = (x-a)(x-b)(x-c)(x-d) - 1$ where $a, b, c, d$ are distinct integers. Prove that $f(x)$ can't be factorized into integer polynomial with $deg ≥1$
In the above question I proved for three degree polynomial using contradiction. But couldn't use the same for the even degree polynomial.If anyone out there could help me would be of great help.

By rational root test or Gauss’s Lemma (see the proof there for the lemma over the integers), if $f(x)$ has linear factor over $\mathbb Q$, then $f(x)$ has a factor of the form $x-t,t\in {\mathbb Z}$. But $f(t)=0$ implies $$(t-a)(t-b)(t-c)(t-d)=1,$$ which is impossible, as $t-a,t-b,t-c,t-d$ are nonzero distinct integers.
If $f(x)$ is reducible but of no linear factor, then by Gauss’s Lemma, $$f(x)=(x-a)(x-b)(x-c)(x-d)-1$$ $$=(x^2+Ax+B)(x^2+Cx+D),$$ where $A,B,C,D\in {\mathbb Z}$. It follows that $$f(a)=(a^2+Aa+B)(a^2+Ca+D)=-1$$
$$\Rightarrow a^2+Aa+B=\pm 1.$$ Similarly $$u^2+Au+B=\pm 1,~{\rm for~}u=b,c,d.$$
By the pigeonhole principle (and permuting $a,b,c,d$, $(A,B)$,$(C,D)$ if necessary), one may assume that $$\left\{\begin{array}{c}a^2+Aa+B=1\\ b^2+Ab+B=1\end{array}\right.\Leftrightarrow \left\{\begin{array}{c}a^2+Ca+D=-1\\ b^2+Cb+D=-1\end{array}\right.,$$
where $A,B,C,D$ can be solved as $$A=-(a+b),B=1+ab,C=-(a+b),D=-1+ab.$$ It follows that $$(x-a)(x-b)(x-c)(x-d)-1$$ $$=(x^2-(a+b)x+(1+ab))(x^2-(a+b)x+(-1+ab)). \quad (1)$$ By comparison of coefficient of $x^3$ and the constant term in (1), one has $$a+b=c+d,abcd-1=a^2b^2-1.\quad (2)$$
Case 1: $ab\neq 0\Rightarrow a+b=c+d,ab=cd,$ which shows that $\{a,b\}=\{c,d\},$ a contradiction to the condition that $a,b,c,d$ are distinct.
Case 2: $ab=0$. By symmetry, it suffices to consider the case $a=0$ (so $bcd\neq 0$). Then (2) gives that $b=c+d$, hence from (1), one has $$x(x-(c+d))(x-c)(x-d)-1$$
$$=(x^2-(c+d)x+1)(x^2-(c+d)x-1).\quad (3)$$ By comparison of coefficient of $x$ in (3), one has $$-cd(c+d)=0,$$ which shows that $c+d=0$ (since $cd\neq 0$). This implies that $$0=c+d=a+b=b,$$ contradiction again. 
One concludes that if $a,b,c,d$ are distinct integers, then $$f(x)=(x-a)(x-b)(x-c)(x-d)-1$$ is irreducible over $\mathbb Q.$ 
