Prove $3\mid σ(3n + 2)$ and $4\mid σ(4n + 3)$ for any positive integer n. In this problem, $\sigma$ is sum of the positive divisors of $n$.
 A: We solve the $4n+3$ problem. Suppose that $4n+3$ has prime power factorization
$$4n+3=p_1^{a_1}p_2^{a_2} \cdots p_k^{a_k}.$$
Since the function $\sigma$ is multiplicative, we have
$$\sigma(4n+3)=\sigma(p_1^{a_1})\sigma(p_2^{a_2}) \cdots \sigma(p_k^{a_k}).$$
The number $4n+3$ must have at least one prime divisor congruent to $3$ modulo $4$ occurring to an odd power. Without loss of generality we may assume that $p_1\equiv 3\pmod{4}$ and that $a_1$ is odd.
We show that $4$ divides $\sigma(p_1^{a_1})$. Note that
$$\sigma(p_1^{a_1})=\frac{p_1^{a_1+1}-1}{p_1-1}.$$
Finally, since $a_1+1$ is even, we have that $p_1^{a_1+1}-1$ is divisible by $8$. Since $2$ is the highest power of $2$ that divides $p_1-1$, the result follows.  Alternately, the sum $1+p_1$ is divisible by $4$, as is the sum $p_1^2+p_1^3$, and so on.
The result for $3n+2$ can be proved in a very similar way.
A: HINT: Suppose that $a\mid 3n+2$. Then there is an integer $b$ such that $ab=3n+2$. Show that one of $a$ and $b$ is congruent to $1$ and the other to $2$ modulo $3$, and conclude that $3\mid a+b$. The other problem is very similar.
