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    Why does this work:

    Take a number, say 4 to keep it simple, and write down it's divisors:

    1,2,4

    Write down the number of divisors these numbers have:

    1,2,3

    Add these numbers and square:

    1+2+3=6
    6 squared = 36

    Now go back and cube the number of divisors of the original divisors:

    1,8,27

    Add these together

    1+8+27=36

    I can see we get 36 twice, but why?

    (Marked as all levels because it's meant to be an investigation for all levels, not just people who do undergrad number theory.)
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    (Original post by rayquaza17)
    Why does this work:

    Take a number, say 4 to keep it simple, and write down it's divisors:

    1,2,4

    Write down the number of divisors these numbers have:

    1,2,3

    Add these numbers and square:

    1+2+3=6
    6 squared = 36

    Now go back and cube the number of divisors of the original number (4):

    1,8,27

    Add these together

    1+8+27=36

    I can see we get 36 twice, but why?

    (Marked as all levels because it's meant to be an investigation for all levels, not just people who do undergrad number theory.)
    Are you sure the bolded and underlines bit is what you meant to say there?
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    (Original post by Zacken)
    Are you sure the bolded and underlines bit is what you meant to say there?
    I think instead of number I was meant to say divisors. Updated OP, sorry!
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    (Original post by rayquaza17)
    I think instead of number I was meant to say divisors. Updated OP, sorry!
    No problem, it's a very cool observation! I can already think of a way into it, will update back with some thoughts.
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    Okay, so right off the bat - standard summation shows us that

    \displaystyle 

\begin{equation*}\sum_{r=1}^n r^3 = \frac{1}{4}n^2(n+1)^2 = \left(\frac{1}{2}n(n+1)\right)^2 = \left(\sum_{r=1}^{n} r\right)^2 \end{equation*}

    Essentially: 1^3 + 2^3 + \cdots + n^3 = (1 + 2 +\cdots +n)^2. That's all fine and dandy but not what we're looking at here. We've proven that the list or set \{1, 2,\cdots n\} has the above property, let's call this property \mathcal{P}.

    Take two sets A and B then it is easy to show that if both of these lists has the property \mathcal{P} then so does their Cartesian product, straight from the definition, we have:

    \displaystyle 

\begin{equation*}\sum_{x \in A \times B} x = \sum_{a \in A} a \sum_{b \in B} b\end{equation*}

    And supposing that both sets have the property, then by the definition:

    \displaystyle 

\begin{equation*}\sum_{a \in A} a^3 = \left(\sum_{a \in a} a\right)^2 \quad \text{and} \quad \sum_{b \in B} b^3 = \left(\sum_{b \in B} b \right)^2\end{equation*}.

    Now multiply both of the above two together, some manipulation and reasoning shows that you that A \times B has property \mathcal{P} and hence (by extension, or induction if you want to be rigorous) if every X_i satisfies \mathcal{P} so does X_1 \times \cdots \times X_n.

    Now consider the integer n = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_k^{\alpha_k} and the rest falls out by considering the number of divisors.
 
 
 
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