Finding integral solutions to the equation $x^4-ax^3-bx^2-cx-d=0$

Question

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.

Answer 1

By the rational root theorem, any integral solution must divide $d$.

Answer 2

By Rational Root Theorem, any integral root will divide d. List out the factors of d, including both 1 and d and considering $\pm$ versions of each. Sub these factors into the polynomial to see which satisfy the equation.

Also, by Descartes' Rule of Signs, you are guaranteed a single positive real root and either 3 or 1 negative real root(s). That might help to narrow down your search, so you can stop more quickly once you've found the requisite number. Note that the Rule of Signs only talks about real roots, not about integral roots specifically.

EDIT: Initially deleted this post, because I wasn't sure what the OP was asking: a general method, or some specific answer for the number of integer solutions of the whole class of degree 4 Diophantine polynomials of that form? I can't see a way to the latter.

Answer 3

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.