I am afraid to continue to ask trivial things but really I do not know how to proceed, so I ask the experts:


A specially multiplicative function, is a function
$f$ from positive integers to complex numbers,
that satisfies:
$$
f(n)f(m) = \sum\_{d \mid \gcd(m,n)} f(\frac{mn}{d^2})g(d)
$$
for all $m,n$, where $g$ is a completely multiplicative function, i;e.
$$
g(rs) = g(r)g(s)
$$
for all positive integers $r,s.$


Seems well known that any specially multiplicative function $F$ is the Dirichlet convolution of $2$
completely multiplicative functions. (We write, say, $F = r \* s$).
That is:


$$
F(n) = r \* s (n) = \sum\_{d \mid n} r(d) s(n/d)
$$
with $r,s$ completely multiplicative functions.


I checked that


$$
\sigma = id \* z
$$


where $\sigma$ is the sum of divisors's function, (that is specially multiplicative with a explicit $g$),
$$
id(n) = n
$$
for all $n$ and
$$
z(n) = 1
$$
for all $n.$


But I am (unfortunately) unable to find completely multiplicative functions $a,b$
such that


$$
\tau = a \* b
$$


where $\tau$ is Ramanujan's tau function. 


Question 1 : Somebody can display such $a,b$ ???


Probably, there is a standard construction of such $a,b$ for any specially multiplicative function.


Question 2 : What is the recipe to get such $a,b$ for a given specially multiplicative function ???



Explicit formula for completely multiplicative functions $a,b$ such that $\tau = a * b$; $\tau$ is Ramanujan's tau function, and $*$ is the Dirichlet convolution.
I don't understand your definition of "specially multiplicative"; if $m$ and $n$ are coprime, it gives $f(m)f(n)=mn$, which is not true of the sum of divisor function...


However, the Ramanujan $\tau$-function satisfies $\tau(m)\tau(n)=\sum\_{d\mid (m,n)}{\mu(d)\tau(mn/d^2)}$, which might be what you mean?


In any case, $\tau$ is indeed the Dirichlet convolution of two totally multiplicative functions: factor the local $p$-factor of the Dirichlet generating series, which is $1-\tau(p)p^{-s}+p^{11-2s}$, as $(1-a(p)p^{11/2-s})(1-b(p)p^{11/2-s})$, then the totally multiplicative functions mapping $p$ to $a(p)$ and $b(p)$ satisfy $a\* b=\tau$. 


There is therefore (uncountably) many possibilities for these functions (with $a(p)$ being chosen out of two possibilities for every $p$), so it is hard to get a simple formula (especially a formula that does not involve factoring $n$ into primes).


