I stumbled across an interesting identity while taking a test for an advanced math tutoring job. I'm pretty busy these days, so maybe somebody else can prove or disprove it before I get around to it in the next couple weeks.

Prove or disprove:

1 + 2^3 + 3^3 + ... + n^3 = [1 + 2 + 3 + ... + n]^2 for all n >= 1

Enjoy!

I'd never seen that one.. it is interesting. A quick search on the internet shows this site with a proof.

http://www.curiousmath.com/index.php?name=News&file=article&sid=59

I checked out the link, and I don't think that's a valid proof (it uses circular logic). They're right about using mathematical induction, but they haven't actually done it yet. So I think the problem is still open. . . .

Well, it's Sunday night and nobody's biting, so here goes. Proof by induction, old-school. First of all, I'll be using a fairly common identity, ∑_1→n i = n(n+1)/2

Let P_n be the statement: ∑_1→n i[sup]3[/sup] = (∑_1→n i)[sup]2[/sup]

Clearly, P₁ is true.

Let m > 1, assume P_m-1 is true.

4m[sup]3[/sup] = 4m[sup]3[/sup]
4m[sup]3[/sup] = m[sup]2[/sup] [big][[/big](2m) - (-2m)[big]][/big]
4m[sup]3[/sup] = m[sup]2[/sup] [big][[/big](m[sup]2[/sup]+2m+1) - (m[sup]2[/sup]-2m+1)[big]][/big]
4m[sup]3[/sup] = m[sup]2[/sup](m+1)[sup]2[/sup] - m[sup]2[/sup](m-1)[sup]2[/sup]
m[sup]3[/sup] = [big][[/big]m(m+1)/2[big]][/big][sup]2[/sup] - [big][[/big](m-1)m/2[big]][/big][sup]2[/sup]
∑₁[sup]m[/sup]i[sup]3[/sup] - ∑₁[sup]m-1[/sup]i[sup]3[/sup] = (∑₁[sup]m[/sup]i)[sup]2[/sup] - (∑₁[sup]m-1[/sup]i)[sup]2[/sup]
∑₁[sup]m[/sup]i[sup]3[/sup] = (∑₁[sup]m[/sup]i)[sup]2[/sup]

Therefore, P_m-1 implies P_m for all m > 1.
And since we already know P₁ is true, P_n must be true for all positive n.