Problem A. 378. (September 2005)
A. 378. Does there exist a function such that f(x)=f(y) whenever x and y are different rationals and xy=1 or x+y{0,1}?
(5 pont)
Deadline expired on 17 October 2005.
Solution. There exists such a function.
Consider the continued fraction form of the rational numbers. As is wellknown, each rational x can be uniquely written as
where a_{0} is an integer, a_{1},...,a_{n} are positive integers and a_{n}>1.
For an arbitrary rational x, lett x be the number of divisions in the continued fraction form of x (the number n above). Then (x)=0, for inegers and (x)=(1/x)+1 for all 0<x<1.
Now define function f. Set
f(0)=1; for all x>0; , for all x<0.
By the definition of (x) we have that for all positive rational x and positive integer k, (x+k)=(x) and f(x+k)=f(x).
Now prove f(1/x)=f(x), if x0,1. Since f(x)=f(x) and f(1/x)=f(1/x), it can assumed that x>0. Moreover, the roles of x and 1/x are symmetric so we can assume 0<x<1. Then (x)=(1/x)+1, therefore f(1/x)=f(x).
From the definition of f we know that f(x)=f(y) for all x+y=0, except x=y=0. The final property which must be verified is that f(1x)=f(x) if . By the symmetry assume .
If x>1 then f(x)=f(x1)=f(1x), done.
If x=1 then f(x)=f(1)=1 and f(1x)=f(0)=1, done.
The only remaining case is when . then
Therefore, function f satisfies all the required properties.
Remark. It is easy to see that there exist exactly two solutions: f and f.
Statistics:
23 students sent a solution.  
5 points:  Erdélyi Márton, Gyenizse Gergő, Hujter Bálint, Jankó Zsuzsanna, KisfaludiBak Sándor, Kónya 495 Gábor, Molnár 999 András, Nagy 224 Csaba, Paulin Roland, Sümegi Károly, Szabó 108 Tamás, Ureczky Bálint. 
4 points:  Korándi Dániel, Radnai András, Szilágyi Dániel. 
3 points:  2 students. 
2 points:  3 students. 
0 point:  3 students. 
