# Prove $a\ge b \ge c \ge d \ge 1 \implies x^{4} -ax^{3}-bx^{2}-cx-d$ has no integer root.

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?

• 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 $x$. Jul 4, 2022 at 8:38

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.
• Why $A \ge C$ implies $x^{2}$ cannot divide $C$? Jul 4, 2022 at 6:01
• @Redsbefall $C \ge x^2$ but $A \ge C$, so $A \ge x^2$ and $Ax^2$ is too big to be equal to $x^4$. Jul 4, 2022 at 6:10
• (+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 $d \le a$.