# $\delta_2(n)$ the sum of the squares of the positive divisors of $n$.

The question is:

Develop a formula for $\delta_2(n)$, the sum of the squares of the positive divisors of $n$.

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$\delta_2$ is usually denoted $\sigma_2$. – lhf Jul 4 '12 at 10:35

Hopefully, this can help. Let's say you wanted to find the sum of the positive divisors of $180=2^2\times3^2\times5$. The sum can be written as

$$(1+2^1+2^2)(1+3^1+3^2)(1+5)$$

Since each of the factors is a geometric series, this can be rewritten as

$$\dfrac{2^3-1}{2-1}\times\dfrac{3^3-1}{3-1}\times\dfrac{5^2-1}{5-1}$$

Can you see how to alter this method for $\delta_2(180)?$

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Thank you. I ended up showing it and arrived at the formula. Here is it, $\delta_2(n)=(\frac{p_1^{2e_1+2}-1}{p_1^{2} - 1})(\frac{p_2^{2e_2+2}-1}{p_2^{2}-1})...(\frac{p_k^{2e_k+2}-1}{p_k^{2}-1})$ :) – HowardRoark Jul 4 '12 at 10:23

Hint: prove it's multiplicative, then evaluate it on prime powers.

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I can prove that it is multiplicative. I have been on this problem for a while now and I finally decided to ask for help. I can't get the formula. – HowardRoark Jul 4 '12 at 9:32
So you can't find $\delta_2(p^k)$ for prime $p$? – Gerry Myerson Jul 4 '12 at 9:47
I got it :) Thank you very much. I thought they meant something else by a formula. I don't know what I was thinking. Thank you, I really appreciate that you gave me the hint instead of the answer. It feels much rewarding that way when do finally arrive at the answer. – HowardRoark Jul 4 '12 at 10:20