<s>[INST] Finding integral solutions to the equation $x^4-ax^3-bx^2-cx-d=0$ How many integral solutions exist for the equation:
$$x^4-ax^3-bx^2-cx-d=0\qquad a\ge b\ge c\ge d\qquad a,b,c,d\in\Bbb{N}$$
I have no idea where to begin even.Please help.
 [/INST] Let's find integer root $x$.
As $d\ne 0$, then $x\ne 0$.
Denote $y=1/x$.
Then equation for $y$:
$$
ay+by^2+cy^3+dy^4=1.\tag{1}
$$
A. If $x=-n$, where $n\in\mathbb{N}$, then $y=-\dfrac{1}{n}$, and $(1) \implies$ 
$$
\dfrac{-a}{n}+\dfrac{b}{n^2}+\dfrac{-c}{n^3}+\dfrac{d}{n^4}=1.\tag{2}
$$
$$
\dfrac{-an+b}{n^2}+\dfrac{-cn+d}{n^4}=1.
$$
LHS is non-positive, since $a\ge b$, $c\ge d$. So, there are no solutions in the form $x=-n, n\in \mathbb{N}$.

B. If $x=n$, where $n\in\mathbb{N}$, then $y=\dfrac{1}{n}$, and $(1) \implies$ 
$$
\dfrac{a}{n}+\dfrac{b}{n^2}+\dfrac{c}{n^3}+\dfrac{d}{n^4}=1.\tag{3}
$$
$(3)$ $\implies$ $n\ne 1$, and
$$
\dfrac{a}{n}<1,\tag{4'}
$$
$$
1\le\dfrac{a}{n}+\dfrac{a}{n^2}+\dfrac{a}{n^3}+\dfrac{a}{n^4}< \sum_{j=1}^{\infty}\dfrac{a}{n^j}=\dfrac{a}{n-1},\tag{4''}
$$
$(4'), (4'') \implies$
$$
\dfrac{n}{a}>1,\tag{5'}
$$
$$
\dfrac{n-1}{a}<1.\tag{5''}
$$
There are no such $n\in \mathbb{N}$ to fit both $(5'), (5'')$ (remember, that $a\in \mathbb{N}$): if $n=a$, then $\dfrac{n}{a}=1$. 
Answer: $0$ integer solutions.
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