Four integers are marked on a circle. On each step we simultaneously replace each number by the difference between this number and 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 $a$, $b$, $c$ and $d$ such that the numbers $|bc-ad|$, $|ac-bd|$ and $|ab-cd|$ are primes?
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Peter
29.08.2007 15:18
No, it's not. Iterating, we consequently get $ (a-b,b-c,c-d,d-a)$, $ (a-2b+c,b-2c+d,c-2d+a,d-2a+b)$, $ (a-3b+3c-d,b-3c+3d-a,c-3d+3a-b,d-3a+3b-c)$, $ (2a-4b+6c-4d,2b-4c+6d-4a,2c-4d+6a-4b,2d-4a+6b-4c)$, ... and we note that the fourth step has only even numbers. Since every next step conserves the factor $ 2$, $ |BC-AD|,|AC-BD|, |AB-CD|$ are all multiples of $ 4$ and thus not prime.
bboypa
06.05.2008 02:37
from the first step $ a + b + c + d = 0$ that we can factorize and obtain that $ |(bc - ad)(ac - bd)(ab - cd)| = ((a + b)(b + c)(c + a))^2$, contradiction.
Pythagorasauras
21.05.2015 12:11
What is the contradiction?
Anajar
14.05.2020 18:24
Pythagorasauras wrote: What is the contradiction? If we suppose that they are all prime we conclude that each of them should be a perfect square. Contradiction. Also if at least two of them are equal we still get contradiction.