by
abacadaea, Feb 18, 2010, 10:15 PM
by
The QuattoMaster 6000, Feb 19, 2010, 12:04 AM
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such that for all 
one has![\[ f(km) + f(kn) - f(k)f(mn) \ge 1.\]](http://latex.artofproblemsolving.com/6/3/b/63bf1a912ab96babfa3c107c3e4b5899c2ad1c85.png)
,
is in the domain as well, so this is harder.)
for all integers 
;
.
to get that 
.
to attain that 
for all integers 
.
so 
.
and 
(so 
)
.
;
,
,
satisfying 
.
so that 
;
gives that
,
for all integers 
;
for all 
.
and ![$ f(m^2) + f(m) - [f(m)]^2\ge1$](http://latex.artofproblemsolving.com/a/f/a/afac711596b388a5385f70eae46da4fcc167b19d.png)
(*). Also set 
so 
.![\[ 3f(m) - [f(m)]^2\ge2\implies1\le f(m)\le2\]](http://latex.artofproblemsolving.com/7/b/c/7bccd0237515eb95659bec2ab84645f62ae79415.png)
for all positive integers ![\[ f(m^2) - f(m)\ge[f(m)]^2 - 2f(m) + 1 = (f(m) - 1)^2,\]](http://latex.artofproblemsolving.com/f/e/f/fef641b89f997d139da81f5c0314b5db452ca7f9.png)
so assume for the sake of contradiction that there is some 
such that 
;![\[ f(t^2) - f(t)\ge(f(t) - 1)^2 = \epsilon^2,\]](http://latex.artofproblemsolving.com/c/b/9/cb9c65ecca2b7e27ea5f96540634602cc3d4dc0f.png)
so 
,![\[ f(t^{2^2}) - f(t^2)\ge(f(t^2) - 1)^2 > \epsilon^2.\]](http://latex.artofproblemsolving.com/d/e/2/de29ec50007de6e9a170dcc6f9b4ddaf14bc0b8e.png)
increases by more than 
for each increase in 
,![\[ f(t^{2^r}) - f(t^{2^{r - 1}}) > \epsilon^2,\]](http://latex.artofproblemsolving.com/e/3/0/e30dc0e698efb429b856012864cf1a0acd4de7c7.png)
so
gives