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If $a,b,c,d$ are positive integers such that $a\ge b\ge c \ge d\ge 1$, prove that

$$ x^{4} -ax^{3}-bx^{2}-cx-d $$

has no integer root.


Attempt:

If there is an integer solution $x=x_{0}$, then

$$x_{0}^{4} = a x_{0}^{3}+ b x_{0}^{2}+c x_{0}+d$$

Notice that it is clear $x_{0} | d$.

Notice that $x_{0} \ne 0$ because $d>0$. If $x_{0}=1$ then, $1=a+b+c+d$, which is impossible because $a+b+c+d \ge 4$. So $x_{0} \notin \{0,1\}$. If $x_{0} = -1$, then $1 = (b-a) + (d-c) \le 0$, contradiction. So $x_{0} \notin \{-1,0,1\}$. So far we can say that $|x_{0}| \ge 2$.

Now if $x_{0} < 0$, then $x_{0}^{4} \ge 16 > 0$. But $a x_{0}^{3},cx_{0} < -1$ with

$$ |a x_{0}^{3} | > b x_{0}^{2} $$ $$ |c x_{0} | > d $$

so we have $x_{0}^{4} = ax_{0}^{3} + bx_{0}^{2} + cx_{0} + d < 0$ contradiction.

So we must have $x_{0} \ge 2$. Now, since $x_{0}| d$ then $d = e x_{0}$, where $e$ positive integer. But this means $ a \ge b \ge c \ge d \ge x_{0}$, which means

$$ax_{0}^{3}+bx_{0}^{2}+cx_{0}+d > x_{0}^{4}$$ contradiction.

Some parts of this proof are not necessary I know, I was just working on it while writing it in this post.

Are the better/more elegant solutions?

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  • $\begingroup$ Integer root $x$ should divide $d$, then $x$ should be less than $d$, but $x=a+b/x+c/x^2+d/x^3$. RHS is greater than $d$ for positive $x$ and is greater or equal than $0$ for negative integer $x$. $\endgroup$ Jul 4, 2022 at 8:38

1 Answer 1

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The case $x=0$ is trivial.

For $x>0$, you have $x^4=(ax+b)x^2+(cx+d)$, with $ax+b\ge cx+d>1$. Let's say that $A=ax+b$ and $C=cx+d$. That means that $x^4=Ax^2+C$ and $C$ must be divisible by $x^2$ but $A\ge C\ge x^2$ so you can't find a solution.

For $x<0$, we have $A\le C\le0$ so $Ax^2+C$ will never be positive. So once again, no solution.

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    $\begingroup$ Why $A \ge C$ implies $x^{2}$ cannot divide $C$? $\endgroup$
    – Redsbefall
    Jul 4, 2022 at 6:01
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    $\begingroup$ @Redsbefall $C \ge x^2$ but $A \ge C$, so $A \ge x^2$ and $Ax^2$ is too big to be equal to $x^4$. $\endgroup$
    – Erick Wong
    Jul 4, 2022 at 6:10
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    $\begingroup$ (+1) Just a nitpick, but since you work with integer $x$ it might be easier to follow if you wrote $x \ge 1$, $x \le -1$ instead of $x \gt 0$, $x \lt 0$. The first part could also be justified by noting that $\,p(a) \lt 0\,$ so the (unique) positive root must be $\gt a$ and therefore cannot divide $d \le a$. $\endgroup$
    – dxiv
    Jul 4, 2022 at 20:13

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