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.
 A: Given that $~f(x) = (x-a)(x-b)(x-c)(x-d) - 1~$ where $~a,~ b,~ c,~ d~$ are distinct integers. 
If possible let, $~f(x)~$ can be factorized into integer polynomial with degree $~\ge 1~$ and let $$~f(x) = (x-a)(x-b)(x-c)(x-d) - 1=p(x)q(x)~$$ where $~p(x)~$ is a linear polynomial and $~q(x)~$ is cubic polynomial.
Then $$p(a)q(a) = p(b)q(b) = p(c)q(c) =p(d)q(d)=−1.$$ If the coefficients of $~p(x)~$ and $~q(x)~$ must integers they can take only integer values, so in each product one of the factor must be $~1~$ and the other one is $~−1~$. Hence either $~p(x)~$ takes the value $~1~$ twice or it takes the value $~−1~$ twice. But a first degree polynomial
cannot take the same value twice.
Hence a  contradiction arises. 
A: 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 $\mathbb Q.$ 
