$a =\prod_{i = 1}^{r} p_{i}^{ai}$ with $a_{i} > 0$ for each $i$ is the canonical representation of $a$, deduce a formula for the sum of the squares of the positive divisors of $a$.
-The section we are currently studying involved purely numerical GCD and LCM questions along with some questions involving finding the number of positive divisors and sum of positive divisors. we did not although go over proofs such as this and I am a bit lost as to how to structure it. Any help is really appreciated.
You can use the same procedure used to find the formula for the sum of the divisors, only that you must consider the squares of the divisors.
For the sum of the divisors of $$ n = \sum_{i=1}^r p_i^{a_i}, $$ which is called $\sigma(n)$, you may reason in this way.
Each divisor will contain the prime $p_i$ to a power between 0 and $a_i$.
For example, if $n=3^2\cdot 7^3$, in each divisor the prime 3 will appear as $3^0$, $3^1$ or $3^2$ and the prime 7 will appear as $7^0$, $7^1$, $7^2$ or $7^3$.
You can image these divisors in a table like $$ \begin{array}{c|c|c} 3^0\cdot 7^0 & 3^1\cdot 7^0& 3^2\cdot 7^0 \\\hline 3^0\cdot 7^1 & 3^1\cdot 7^1& 3^2\cdot 7^1 \\\hline 3^0\cdot 7^2 & 3^1\cdot 7^2& 3^2\cdot 7^2 \\\hline 3^0\cdot 7^3 & 3^1\cdot 7^3& 3^2\cdot 7^3 \end{array} $$
It's easy to compute the sum column by column. In the first row you can collect $3^0$ and obtain $3^0\cdot (7^0+\cdots+ 7^3)$. The same for the second column $3^1\cdot (7^0+\cdots + 7^3)$ and the third $3^2\cdot (7^0+\cdots + 7^3)$.
Since the last term of these products is the same, you can collect it, and you obtain that the sum of the divisors of $n$ is $$ (3^0+3^1+3^2)\cdot (7^0+7^1+7^2+7^3) $$
At this point it is easy to generalize if $n$ is the product of the powers of 3 primes. The sum of the divisors will be $$ (p_1^0+p_1^1+\cdots+p_1^{a_1})(p_2^0+p_2^1+\cdots+p_2^{a_2})(p_3^0+p_3^1+\cdots+p_3^{a_3}) $$
From this we can arrive to the general case in which there are $r$ prime factors.
The sum of the divisors will be equal to the product of all the terms of the form $p_j^0+p_j^1+\cdot+p_j^{a_j}$ for $j$ from 1 to $r$. In general the sum $x^0+x^1+\cdot+x^k$ is equal to $(x^{k+1}-1)/(x-1)$ for $x\neq1$, so this can be rewritten as $$ \sigma(n) = \prod_{i=1}^r\frac{p_i^{a_i+1}-1}{p_i-1} $$
This is the general idea. By following this blueprint you should be able to find easily the sum of the squares of the divisors.