TITLE: There are no integral roots of $x^4-ax^3-bx^2-cx-d=0,$ given $a>b>c>d; \;\;a,b,c,d \in \mathbb N$
QUESTION [9 upvotes]: Prove that roots of the equation $x^4-ax^3-bx^2-cx-d=0$ are non integer. Where $a>b>c>d; \{a,b,c,d\}\subset \mathbb{N}$.
My progress :
Let $f(x)=x^4-ax^3-bx^2-cx-d$
$\implies f(0)=-d<0,f(-1)=1+a-b+c-d>0$
$\implies$ at least one root $\in (-1,0)$
$ \implies$ at least two non integer roots as sum of all roots is an integer.

REPLY [3 votes]: Let $x=-m$ be a root, $m\in\mathbb{N}$
$\implies m^4+m^2(am-b)+(cm-d)=0$ which is a contradiction as each parenthesis is positive.
Obviously root can’t be $0$
Let $x=m$ be a root, $m\in\mathbb{N}$
$\implies m^4=am^3+bm^2+cm+d$ 
$\implies m|d\implies d=km\implies a=km+s,s\in \mathbb{N}$
$\implies m^3=km^3+(s+b)m^2+c+k$
which is a contradiction as $km^3\ge m^3$ and rest of the term of RHS are positive.
Hence given equation has no integral roots !!