IMO 1996 SL N1

Origin: UKR | Category: Number Theory

Problem

Four integers are marked on a circle. At each step we simultaneously replace each number by the difference between this number and the next number on the circle, in a given direction (that is, the numbers $a,b,c,d$ are replaced by $a-b$, $b-c$, $c-d$, $d-a$). Is it possible after $1996$ such steps to have numbers $a,b,c,d$ such that the numbers

$$|bc-ad|,\quad |ac-bd|,\quad |ab-cd|$$

are primes?

Solution

It is easy to check that after $4$ steps we will have all $a,b,c,d$ even. Thus

$$|ab-cd|,\quad |ac-bd|,\quad |ad-bc|$$

remain divisible by $4$, and clearly are not prime. The answer is no.

Second solution

After one step we have

$$a+b+c+d=0.$$

Then

$$ac-bd=ac+b(a+b+c)=(a+b)(b+c),$$

etc., so

$$|ab-cd|\cdot|ac-bd|\cdot|ad-bc| =(a+b)^2(a+c)^2(b+c)^2.$$

However, the product of three primes cannot be a square, hence the answer is no.