A function $f$ is defined from a triplet of positive reals to a positive real number $f:R_+^3 \to R_+$ and satisfies the following:

1. $f$ is symmetric in all 3 variables. That is $f(a,b,c) = f(b,a,c) = f(a,c,b) = ...$

2. For any positive real $t$, $f(ta,tb,tc) = tf(a,b,c)$

3. $f(1/a, 1/b, 1/c) = 2011^2/f(a,b,c)$

4. $f(a,b,c) = f(\sqrt{ab}, \sqrt{ab}, c)$

Determine how many triplet of positive integers $(x,y,z)$ are there such that $f(1/x,1/y,1/z) = 1$

SolutionFrom rule 3, we have $f(1,1,1) = 2011^2/f(1,1,1)$, so $f(1,1,1) = 2011$.

Now, for any $u,v > 0$,

$f(u,v,1/(uv)) = f(\sqrt{uv}, \sqrt{uv}, 1/(uv))$ using rule 4

and

$f(uv,1,1/(uv)) = f(\sqrt{uv}, \sqrt{uv}, 1/(uv))$ using rule 4

So

$f(u,v,1/(uv)) = f(uv,1,1/(uv)) = f(uv, 1/(uv), 1) = f(1,1,1) = 2011$

Now, for any $t,u,v$, let $k = \sqrt[3]{tuv}$, then $t = k^3/uv$

$f(t,u,v) = f(k^3/uv, u, v) = kf(k^2/uv, u/k, v/k) = kf(u/k, v/k, 1/((u/k)(v/k))) = 2011k = 2011\sqrt[3]{tuv}$

Thus $f(t,u,v) = 2011\sqrt[3]{tuv}$ for all $t,u,v$.

Now we need to determine the number of positive integer triples $(x,y,z)$ such that

$$f(1/x,1/y,1/z) = \frac{2011}{\sqrt[3]{xyz}} = 1$$

$$xyz = 2011^3$$

Since 2011 is a prime, there are only the following possibilities:

1. $(1,1,2011^3)$ and all its permutations, there are three triplets.

2. $(1,2011, 2011^2)$ and all its permutations, there are six triplets.

3. $(2011,2011,2011)$, there is one triplet.

So in total, there are ten triplets that satisfy the condition.

The above conditions could be tailored to fit any symmetric functions. For example, for $f = a^2 + b^2 + c^2 + k$ one can have the following conditions:

1. $f$ is symmetric

2. $f(\sqrt{a^2+t},b,c) = f(a,b,c) + t$

3. Any condition that would determine $f(0,0,0)$

4. $f(a,b,c) = f(\sqrt{(a^2+b^2)/2},\sqrt{(a^2+b^2)/2},c)$

Likewise, for $f = ab+bc+ca$ one can have the following conditions (really hard):

1. $f$ is symmetric

2. $f(a+t,b,c) = f(a,b,c) + t(b+c)$

3. Any condition that would determine $f(0,0,0)$

4. $f(a,b,c) = f(k,k,c)$ where $k = \sqrt{(a+c)(b+c)}-c$