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

Home > Just Math > Calculus
An increasing function equation (Posted on 2021-01-24) Difficulty: 3 of 5
Find all strictly increasing functions f:N->N such that (f(x) + f(y))/(1 + f(x + y)) is a non-zero natural number, for all x, y∈N.

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution Comment 2 of 2 |
f(x) and f(y) are positive, so 0 < (f(x) + f(y))/(1 + f(x+y))

x and y are positive so x+y is larger than x and y individually, and since f() is strictly increasing then f(x+y) is greater than f(x) and f(y) individually.

Then (f(x) + f(y))/(1 + f(x + y)) < 2*(f(x+y))/(1 + f(x+y))

Decreasing the denominator by 1 will increase the value of the fraction, then 2*(f(x+y))/(1 + f(x+y)) < 2*(f(x+y))/f(x+y) = 2.

Combining all the inequalities gives 0 < (f(x) + f(y))/(1 + f(x+y)) < 2.

But (f(x) + f(y))/(1 + f(x+y)) must be an integer.  The only integer satisfying the inequality is 1, therefore (f(x) + f(y))/(1 + f(x+y)) = 1.

(f(x) + f(y))/(1 + f(x+y)) = 1 can be rearranged into f(x) + f(y) = 1 + f(x+y).

Let x=z+1 and y=z+1.  Then 2*f(z+1) = 1 + f(2z+2)
Let x=z and y=z+2. Then f(z) + f(z+2) = 1 + f(2z+2)

Combining the last two equations yields f(z) + f(z+2) = 2*f(z+1). 

Rearrange this into f(z+2)-f(z+1) = f(z+1)-f(z).  With z, z+1, and z+2 being consecutive integers, this implies that f(z), f(z+1), and f(z+2) are colinear.

Thus f(x) is a linear function.  Let f(x) = mx+b.  Then f(x) + f(y) = 1 + f(x+y) becomes (mx+b) + (my+b) = 1 + m*(x+y) + b.  

All terms with 'm' cancel each other out leaving 2b=b+1, which makes b=1.  Therefore f(x) = m*x+1 for positive integers m.

  Posted by Brian Smith on 2021-01-24 20:54:16
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 (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2021 by Animus Pactum Consulting. All rights reserved. Privacy Information