Does general case of following function exist?

Question

Suppose we have the following: $m_1 m_2 m_3$

Where $m_1$, $m_2$, $m_3$ are constants $> 0$

I'm looking for a function $f$ such that:

$m_1 m_2 m_3 = 1.0 + f(m_1,m_2,m_3) + f(m_2,m_1,m_3) + f(m_3,m_1,m_2)$

and $f(1,x,y) = 0$
and $f(x,1,1) = x-1$
and $f(x,w,z) = f(x,z,w)$

I can't find case for $3$ constants $m_1$, $m_2$, $m_3$. The case for $2$ constants I get following function:

$f(x,y) = [(x-1.0)+(xy-y)]/2$

we have
$1 + f(x,y) + f(y,x)$
$=1+ [(x-1.0)+(xy-y)+(y-1.0)+(yx-x)]/2$
$=2/2 + (2xy-2)/2$
$=xy$

also $f(1,y) = 0.0$
and $f(x,1) = x-1$

Is there a solution for $3$ variables? or general case of $n$ variables?

Answer 1

After a bit of playing around I found a solution for $3$ variables: $$f_3(x,y,z)=\frac{1}{6}\Bigl(2xyz + (xy+ xz -2yz) + (2x -y-z)-2\Bigr).$$ When you check the conditions you see that when summing up the three values with the variables permuted in a cyclic way, everything cancels out except for the $xyz$ terms. Using the same trick you can then easily write down a simple solution for $4$ variables: $$f_4(x,y,z,w)=\frac{1}{12}\Bigl(3xyzw+(xyz+xyw+xzw-3yzw)+(xy+xz+xw-yz-yw-zw)+(3x-y-z-w)-3\Bigr).$$ Generalising to $n$ variables should be then fairly straightforward.