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We have $$ 1^3 + \dotsb + n^3 = (1 + \dotsb + n)^2 $$ as we can establish by induction. But why does this hold? Can we connect it to something else?

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marked as duplicate by Grigory M, egreg, user85798, Dan Rust, Andrew D. Hwang Apr 11 '14 at 23:42

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    $\begingroup$ See my answer here. $\endgroup$ – J.R. Apr 11 '14 at 19:27
  • $\begingroup$ @YourAdHere Is there a higher dimensional generalization of this identity ? $\endgroup$ – Amr Apr 11 '14 at 19:31
  • $\begingroup$ @Amr In case you mean for a higher exponent than $3$, I don't think so, there is a general formula to sum the first $k$th powers and it doesn't take such a simple form for higher $k$. $\endgroup$ – J.R. Apr 11 '14 at 19:35
  • $\begingroup$ @YourAdHere Yes I meant higher exponents. I said "dimensional" because your proof considers squares and I was asking if there is a similar argument that considers cubes for example ... $\endgroup$ – Amr Apr 11 '14 at 19:39
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    $\begingroup$ Related to this question $\endgroup$ – robjohn Apr 11 '14 at 20:22
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Meanwhile, it generalizes to Liouville's $$ \sum_{k | n} \left( d(k) \right)^3 = \left( \sum_{k | n} d(k) \right)^2 $$

Here $d(k)$ is the number of divisors of a positive integer, with $d(1)=1.$ For a prime $p,$ we get $$ d(p^w) = w+1. $$

The identity works because it works for a prime power, that is what the original summation formula shows. Next, both sides are number theoretic "multiplicative." A multiplicative function $f(n)$ is one that applies to integers, and which has this condition: whenever $\gcd(a,b) = 1,$ we have $f(ab) = f(a) f(b).$ Any multiplicative function is completely determined by its values on prime powers. Oh, if $f(n)$ is a multiplicative function, then $$ g(n) = \sum_{k|n} f(k) $$ is also multiplicative. That requires a little proof, double sum sort of thing.

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    $\begingroup$ Fantastic! I had never seen this identity of Liouville. $\endgroup$ – Bruno Joyal Apr 11 '14 at 20:49
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Funnily, we also have

$$\int_0^x t^3\: dt = \left(\int_0^x t\: dt\right)^2.$$

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  • $\begingroup$ +1 But let's see the proof! $\endgroup$ – Aaron Hall Apr 11 '14 at 20:58
  • $\begingroup$ @AaronHall The funny thing is that the integrals $\int_0^x t^n$ were first calculated using Faulhaber polynomials. It's quite possible that the identity of the OP was known before the value of the integral $\int_0^x t^3$! $\endgroup$ – Bruno Joyal Apr 11 '14 at 21:00
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    $\begingroup$ The summation version goes back to antiquity, but integration is new, and I have no idea what Faulhaber polynomials are. They sound impressive. pops corn $\endgroup$ – Aaron Hall Apr 11 '14 at 21:01
  • $\begingroup$ @AaronHall en.wikipedia.org/wiki/… :) $\endgroup$ – Bruno Joyal Apr 11 '14 at 21:02
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There's a famous proof by C. Wheatstone. http://en.wikipedia.org/wiki/Squared_triangular_number

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