$n^2(n-1)\sigma(n)=0 \mod 12$, where $\sigma(n)$ is the sum of divisors function I would like to ask about the following question:

1) Clarify and complete a proof (by cases) in which the multiplicative function $\sigma(n)$, this is the sum of divisors function, satisfies $$n^2(n-1)\sigma(n)=0 \pmod{12}.$$ 
2) Find if it is possible another significatives couples $(P(x), m)$, where $P$ is a polynomial, with integer coefficients and $m\gt 1$ an integer, such that for every $n\geq 1$ 
  $$P(n)\sigma(n)\equiv 0\pmod{m}.$$
To obtain this goal perhaps you need use a computer to conjecture for easy polynomials $P$ and low integers $m$. 

In my case I have not heuristics to ask if there is finitely many of theses cases.
My attempt of solution:
First we have the Fundamental Theorem of Arithmetic and the fact that $\sigma(n)$ is a multiplicative function. Let $n=p^a$, with $p$ an odd prime and $a>1$ an integer (the cases $n=p=2, 3$ are directly deduced, the case $n=2^k$ is easy from Little Fermat and propierties of congruences), then 
$p^2a(p^a-1)\frac{p^{a+1}-1}{p-1}\equiv 0\mod 3$ since $p=2\lambda +1$ is odd, and $p^a-1\equiv 0\mod 4$, then by Binomial theorem $(2\lambda+1)^a-1\equiv \binom{n}{0}+\binom{n}{1}(2\lambda)+\cdots +\binom{n}{n}(2\lambda)^n$. Assume $3\nmid p$ (isn't we finished) then by Little Fermat Theorem $p^2a\equiv 1\mod 3$ thus we test the first factor. 
If $p=3\lambda+1$ by Binomial Theorem we finish.
If $p=3\lambda+2$ we put $a=3$ and we see that $3\nmid 124$, thus we should work with the third factor $1+p+\cdots+p^a=\frac{p^{a+1}-1}{p-1}$; since $p-1=3\lambda+1$ divides $p^{a+1}-1$, and $gcd(3,p-1)=1$ then I work as goal to obtain that $3$ divides $p^{a+1}-1$, but it is $(3\lambda+2)^{a+1}-1$ and Binomial and Little Fermat theorem works.
Remark. I am inspired on the fact that the numbers $\sigma(k)$ satisfy
$$\frac{n^2(n-1)}{12}\sigma(n)=\sum_{k=1}^{n-1}[5k(n-k)-n^2]\sigma(k)\sigma(n-k)$$
a recursion relation derived in [1], and used after by Touchard to prove its theorem about odd perfect numbers. The congruence that we derive from this, when we divide by $12$ is useful for discard some cases in problems involving integers and the sum of divisors function. Really I don't know this result in literature and I belive that is non trivial because in my cases I use factors involving too factor that appears in the sum of divisor function. 

My goal is to finish the proof, any help or hint is appreciated. Thanks in advance.

References (the technics and result aren't significatives to my question, I have not a open access, the relation too appears in a paper of Holdener in the Monthly 109):
[1] Balth. Van der Pol, On a Non-linear Partial Differential Equation satisfied by the Logarithm of the Jacobian Theta Functions, with Arithmetical Applications I, II, Nederl. Akad. Wetensch. Proc., Ser. A, v. 54, 1951, p. 261-271.
 A: 1) We need to show that $n^2(n-1)\sigma(n)$ is divisible both by $4$ and by $3$. 
We have divisibility by $3$ immediately if $3\mid n$ or $3\mid n-1$. In the remaining case $n\equiv 2\pmod 3$ there must be a prime $p\mid n$ with $p\equiv 2\pmod 3$ (as the product of primes $\equiv 1\pmod 3$ is itself $\equiv 1\pmod 3$). Moreover one such prime must occur to an odd power $p^{2k+1}$ (again because even powers are $\equiv 1\pmod 3$). Such a prime power contributes a factor $1+2+p^2+\ldots+p^{2k+1}\equiv 1-1+1-1\pm\ldots-1\equiv 0\pmod 3$ to $\sigma(n)$
We have divisibility by $4$ immediately if $2\mid n$ or $4\mid n-1$. Remains the case $n\equiv 3\pmod 4$. Similar as above, there must exist a prime $p\equiv -1\pmod 4$ that occurs to an odd power in $n$ and contributes an even factor to $\sigma(n)$.
2) First of all we should restrict orselves to primitive polynomials, i.e., the gcd of the coefficients is $1$. If we consider $P(x)=x(x-1)(x-2)\cdots(x-k+1)$, then $k!\mid P(n)$ for all $n$, so if $m\mid k!$ this gives us an example that does not even "use" $\sigma(n)$, so this is not interesting either. The question thus becomes: Can we find primitive $P$ with $m\nmid (\deg P)!$ and what is the minimal possible degree of such $P$? 
