# $1^3 + \dotsb + n^3 = (1 + \dotsb + n)^2$: reason? [duplicate]

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?

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

• See my answer here. – J.R. Apr 11 '14 at 19:27
• @YourAdHere Is there a higher dimensional generalization of this identity ? – Amr Apr 11 '14 at 19:31
• @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 $k$. – J.R. Apr 11 '14 at 19:35
• @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 ... – Amr Apr 11 '14 at 19:39
• Related to this question – robjohn Apr 11 '14 at 20:22

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.

• Fantastic! I had never seen this identity of Liouville. – Bruno Joyal Apr 11 '14 at 20:49

Funnily, we also have

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

• +1 But let's see the proof! – Aaron Hall Apr 11 '14 at 20:58
• @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 $\int_0^x t^3$! – Bruno Joyal Apr 11 '14 at 21:00
• 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 – Aaron Hall Apr 11 '14 at 21:01
• @AaronHall en.wikipedia.org/wiki/… :) – Bruno Joyal Apr 11 '14 at 21:02

There's a famous proof by C. Wheatstone. http://en.wikipedia.org/wiki/Squared_triangular_number