1.

First prove that this is true for n=1:
The factors of $n=1$ are just $1$, so $a=1$, $b=1$, and $c=1$, and we see
that $a \cdot b = 1$.

Prove by well-ordering that for $n \geq 2$, $a \cdot b = c$:

Let $S = \{ n \mid a \cdot b \neq c and where n has no factors that are squares (except 1) \}$.
Assuming $S$ is not empty, then by well-ordering, we must have a least element 
$x \??? S$.  

By the Fundamental Theorem of Arithmetic, $x$ can be factored into a product of 
primes.  Let $p$ be a prime factor of x (which exists since $x \geq 2$), 
and set $m = x/p$.  Of couse, $m < x$.  

Let $a'$, $b'$, and $c'$ be the sum of factors, their squares and
the sum of factors of $m^3$, respectively.  

Since $a'$ is the sum of all factors of $m = x/p$ and $p$ is not a 
prime factor of $m$ (since $x$ has no factors which are
squares, each prime factor appears only once in $x$), $a = a' + p \cdot a'$.
(All the factors of $m$ and all new factors that can be made by multiplying
the old factors by $p$.)

Similarly, $b = b'+ p \cdot b'$.

Now, $x^3 = m^3 p^3$, so to get all the factors of $x^3$, we need all factors
of $m^3$ (which is $c$) and all new combinations that can be made
with $p^3$.  This gives us 
\( c = c' + p c' + p^2 c' + p^3 c'$ = c' (1 + p + p^2 + p^3) \).
Substituting for $a$, $b$, $c$ in $a \cdot b \neq c$, we get 
\( a'(1+p) \cdot b' (1+p^2) \neq c' ( 1 + p + p^2 + p^3) \)
which gives us $a' \cdot b' \neq c'$.  

So m must be in the set $S$ as well, but $x$ is the least element of $S$,
and $m < x$ so we have a constradiction, and $S$ must be empty.