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

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.

Answer 1

You can also use the Rational Root Theorem.

The given monic polynomial has integer roots that are factors of $d$.

Let $q$ be a factor of $d$.

$\dfrac{d}{q}= \lambda \implies q= \dfrac{d}{\lambda}$ $\space{\lambda \in \mathbb{Z}}$

If $q$ satisfies the equation:

$x^4= ax^3+ bx^2 + cx +d$

then:

$\dfrac {d^4}{ \lambda^4} = a\dfrac{d^3}{\lambda^3}+b\dfrac{d^2}{\lambda^2}+c\dfrac{d}{\lambda}+d \implies \dfrac{d^4}{\lambda^4}= \dfrac{ad^3+bd^2\lambda + cd\lambda^2+d\lambda^3}{\lambda^3}\\ \implies \color{blue}{d^4} = \color{red}{ad^3\lambda} +bd^2\lambda^2 + cd\lambda^3 +d\lambda^4$

But

$a>d \implies ad^3 >d^4 \implies ad^3\lambda> d^4$

Hence equality never holds, which is a contradiction, that is, $q$ is not a root.

Hence, the equation has no integral roots.

Edit:

For $\lambda \in \mathbb{Z^{-}}$ Now, this is only possible when $q<0$ Let $q= -m$ , where $m\in \mathbb{Z^+}$

$q = (-m)$ should satisfy the equation:

$m^4 = -am^3+bm^2 -cm +d$

But, $am^3 >bm^2$ and $cm>d$

$\implies$ RHS of the equation is negative, which is again a contradiction since $x^4 > 0$ (here it can't be zero as that would lead to $d=0$),

Hence $\lambda$ cannot be negative.

Answer 2

It might be better to argue the contrapositive: if $n\in\mathbb{Z}$ is a root of $x^4-ax^3-bx^2-cx-d=0$ then it is impossible to have $a\gt b\gt c\gt d\gt 0$. To see this, start by supposing $n\gt 0$. Then the equation is equivalent to $n^4=an^3+bn^2+cn+d$. Now, imagine thinking of this in positional notation: $10000_n=abcd_n$. This suggests breaking into the cases $a\geq n$ and $0\lt a\lt n$. Can you find a contradiction in each of these cases? Once you've figured this out, the $n\lt 0$ case is slightly more complicated but can be handled in very similar fashion.

Answer 3

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 !!