Competition problem — Algebra — Maths Olympiad Prep

Library / /9 of 20

Algebra Difficulty 4.8 AIME Prove it United States

Problem:
Suppose $a, b, c$ are positive integers such that
$\begin{aligned} & b=a^{2}-a \\ & c=b^{2}-b \\ & a=c^{2}-c \end{aligned}$
Prove that $a=b=c=2$ .

Solution

Solution:
If $a=1$ , we get $b=0$ which is impossible. So it is enough to show that $a$ cannot be greater than $2$ . If $a>2$ , we have
$b=a^{2}-a=a(a-1)>a(2-1)=a.$
So $b>a$ ; in particular $b>2$ , so applying the same logic to the second equation we get $c>b$ . Lastly, we have $c>2$ so applying the same logic to the third equation we get $a>c$ . We have now proved $a>c>b>a$ which is a contradiction.

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.