This is further simplified by changing coordinates, i.e. by using a different basis than $n^j$ for the vector space of polynomials. A better basis is $n, n(n-1), n(n-1)(n-2), \dots, n(n-1)\cdots(n-k)$ since these functions have nice sums. In other words, try to find an explicit formula for $\sum_{a=0}^n a(a-1)(a-2)$ and be delighted

Jörg Bewersdorff. Galois Theory for Beginners: A Historical Perspective
http://tinyurl.com/2u6k52

Lagrange came up with the sum-of-roots-of-unity ansatz while trying to understand what’s going on with the quintic.

In fact, I find the usual “modern” presentation of Galois theory utterly incomprehensible. I mean, it’s not so bad and quite simple after knowing the historical approach of permuting the roots. But why, why is the easy way pretty much forgotten? Why struggle months with something abstract-nonsensical and void of motivation when there’s a dead simple and even more illuminating approach to understand it in hours? As Feynman (?) put it: “If you can’t explain it to a six year old, you don’t really understand it.” But I’m preaching to the choir

Let s = a + b + c, t = a + wb +w^2c, u = a + w^2b + wc

Then express t^3 + u^3 and t^3 * u^3 in terms of the symmetric terms.

There is a nice story to that solution you can find it in Mark Kac’s autobiography “Enigmas of Chance”.

It seems to me that this solution is related to Terence Tao’s solution as a sort of inverse transformation.

Since someone cited sums of powers I remembered finding a nice “geometrical” solution to the sum of squares 1^2 + 2^2 + 3^2 +…+ n^2. Organize one 1, two 2’s, three 3’s, …., and n n’s into an equilateral triangle in the natural way, then add the corresponding elements of the three versions of this triangle (the original and the rotated by 120 and 240 degrees) it nice to see that the sum is constant and equal to 2n+1. Since there are n(n+1)/2 elements in the triangle the total sum is n(n+1)/2*(2n+1) and the sum of the elements in the original triangle n(n+1)(2n+1)/6 which is the sum of the squares.

I would like to present this in a motivated way but I am afraid that is somewhat longer. I also tried to generalize this idea to get the sum of cubes but failed. Since it has such a nice sum I suppose that it could also have a nice geometrical method to it. Anyone has an idea?

On an unrelated note, when you say “we end up needing to solve quadratics and take cube roots, both of which we are allowed to assume that we can do”, I wonder how many people would actually consider cube roots of a complex number something basic they “can calculate”, as, unlike in the square root case, you can’t find the real and imaginary parts in terms of roots of only real numbers, so presumably would resort to using transcendental functions (in which case you could of course solve the quintic too.) Utlimately you have to make a choice of “numbers I can deal with” - something you’ve talked about before in other places.

There is a discussion in section 6.5 of Concrete Mathematics, by Graham, Knuth and Patashnik, of how one might discover and prove the formula for this sum by nothing but sheer obstinancy. Later, in section 7.3, they describe how the problem becomes easier when armed with the tool of generating functions. I’ll sketch the latter attack here, because generating functions are a great tool. Set S(m,n)=1+2^m+3^m+..+n^m. There are four approaches you might try in a generating function attack: you could define any of the four functions

A_m(w)=sum_n S(m,n) w^n, B_m(w)=sum_n S(m,n) w^n/n!, C_n(z)=sum_m S(m,n) z^m, or D_n(z)=sum_m S(m,n) z^m/m!

and try to get this function into a simple form. With A, B and C, we strike out. But, if we come back for a fourth swing, D_n has the nice closed form

(e^{nz}-1)/(e^z-1)=(sum_{k>=1} n^k z^k/k! ) (1/z-1/2+z/6-z^3/30+…

and we get the closed formula immediately. So one approach to this sum comes down to knowing about generating functions, being willing to try again when the first attack fails, and some basic comfort manipulating series.