Monday, July 13, 2015
Is there an easier proof? A less messy proof?
Consider the following statement:
For all a,b,c, the equations
x + y + z = a
x2 +y2 + z2 = b
x3 + y3 + z3 = c
has a unique solution (up to perms of x,y,z).
One can also look at this with k equations, k variables, and powers 1,2,...,k.
The STATEMENT is true. One can use Newton's identities (see here) to obtain from the sums-of-powers all of the symmetric functions of x,y,z (uniquely). One can then form a polynomial which, in the k=3 case, is
W^3 -(x+y+z)W^2 + (xy+xz+yz)W - xyz = 0
whose roots are what we seek.
I want to prove an easier theorem in an easier way that avoids using Newton's identities. Here is what I want to prove:
Given those equations above (or the version with k-powers), and told that a,b,c are nonzero natural numbers, I want to prove that there is at most one natural-number solution for (x,y,z) (OR for x1,...,xk in the k-power case).
Its hard to say `I want an easier proof' when the proof at hand really isn't that hard. And I don't want to say I want an `elementary' proof- I just want to avoid the messiness of Newton's identities. I doubt I can formalize what I want but, as Potter Stewart said, I'll know it when I see it.