Competition problem — Algebra — Maths Olympiad Prep

Library / /4 of 121

Algebra Difficulty 4.6 AIME Prove it India

Problem:

Let $a, b, c, d$ be positive integers such that $a \geq b \geq c \geq d$ . Prove that the equation $x^{4}-a x^{3}-b x^{2}-c x-d=0$ has no integer solution.

Solution

Solution:

Suppose that $m$ is an integer root of $x^{4}-a x^{3}-b x^{2}-c x-d=0$ . As $d \neq 0$ , we have $m \neq 0$ .

Suppose now that $m>0$ . Then $m^{4}-a m^{3}=b m^{2}+c m+d>0$ and hence $m>a \geq d$ . On the other hand $d=m\left(m^{3}-a m^{2}-b m-c\right)$ and hence $m$ divides $d$ , so $m \leq d$ , a contradiction.

If $m<0$ , then writing $n=-m>0$ we have $n^{4}+a n^{3}-b n^{2}+c n-d=n^{4}+n^{2}(a n-b)+(c n-d)>0$ , a contradiction.

This proves that the given polynomial has no integer roots.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.