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Number Theory Problem

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.)

(edited 8 years ago)

Reply 1

Zacken

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?

Reply 2

rayquaza17

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!

Reply 3

Zacken

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. :biggrin:

Reply 4

Zacken

Okay, so right off the bat - standard summation shows us that

Unparseable latex formula:

\displaystyle [br]\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:

Unparseable latex formula:

\displaystyle [br]\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:

Unparseable latex formula:

\displaystyle [br]\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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