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such that the equation
be a positive integer and 
be a polynomial with integer coefficients such that 
.
is divisible by 
.
such that for every positive real 
, there exists an integer 
satisfying![\[
\frac{\varphi(n!)}{n^\alpha\,(n-1)!} \;>\; c.
\]](http://latex.artofproblemsolving.com/4/4/4/4443cf7166cc65e67e2e504a826c4ac5f91aab2b.png)
for 
, while 
or 
.
. Note that if 
then 
and 
yield 
, i.e., 
is injective. Next, 
yields![\[
f(x+f(f(y)) + f(f(z))) = z+f(f(y)) + f(f(x)).
\]](http://latex.artofproblemsolving.com/0/b/3/0b39257b2ab3c4794f217b29203253a31c8beef7.png)
Switching the order of 
and using the injectivity of , we find that![\[
x+f(f(y)) = y+f(f(x))\Rightarrow f(f(x)) = x+f(f(1)) - 1.
\]](http://latex.artofproblemsolving.com/a/9/c/a9cf9279116711d166cba77a395e8be18b2fbe26.png)
So, 
gives![\[
f(x+y+z+2f(f(1)) -2) = x+y+z+2f(f(1)) -2\Rightarrow f(x+2f(f(1))) = x+2f(f(1)).
\]](http://latex.artofproblemsolving.com/5/e/a/5ea6db5a9c94c67cfc486984a29daffaaada888d.png)
So, for all 
we find 
. This together with 
gives 
. So, 
for 
. Lastly, implies 
, yielding the answers above.
and
To check that these work, note that 
so we just need to check that 
, which we can see is true for these two functions.
, we get:
Changing 
and 
and comparing, we have that 
for some constant 
. Then the above simplifies to:
or 
. For such 
, from we get 
so 
. Finally, 
, so is bijective.
. If 
then 
, so since is bijective we need 
, leading to the first solution.
then since we have 
, leading to the second solution.
then 
so , contradiction.
,![\[\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4+\frac{8(a-c)^2}{(a+b+c+d)^2}.\]](http://latex.artofproblemsolving.com/7/b/6/7b6e32d32d89ca058a06974e7276f0a5874d90a7.png)
be the set of all positive integers. Find, with proof, all functions 
with the property that 
for all positive integers 
.