All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Cyclic recursive summation (Posted on 2021-07-16) Difficulty: 5 of 5
Let N be the set of positive integers. Find all functions f:N->N that satisfy the equation

fabc-a(abc) + fabc-b(abc) + fabc-c(abc) = a + b + c

for all a, b, c ≥ 2.

(Here f1(n) = f(n) and fk(n) = f(fk-1(n)) for every integer k greater than 1)
(Also note: abc is the product a·b·c and not the concatenation 100a+10b+c)

No Solution Yet Submitted by Danish Ahmed Khan    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
No fixed point for arguments of interest | Comment 6 of 9 |
Define arguments of interest as the product of at least three primes. Note that for ay integer, k, that is not of interest, f(k) is free to take on any value without affecting the validity of the specifying equation.

Let K = abc be an argument of interest and assume that f(abc) = abc. Then abc = a + b + c, which is not true for integers a,b,c > 1. Therefore f has no fixed points among arguments of interest.

  Posted by FrankM on 2021-08-16 22:06:11
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information