Following an additive proof
 

I started reading Nathanson's [i]Additive Number Theory[/i] and I hate to admit that I'm stuck on page 6. The proof is of Lagrange's Four-Square theorem (of course), and here's the setup toward showing that p can be represented as the sum of four squares.

It has just been proven that there is some integer n such that a^2 + b^2 + 1 = np, with 0 < n < p. The proof continues, "Let m be the least positive integer such that mp is the sum of four squares... we must show that m = 1."

How do we know that such a number exists? I mean, without assuming the four-square theorem itself. :razz:

And no, the proof doesn't say -- the RAA proof that m = 1 starts "Suppose not. Then 1 < m < p.", which also assumes the existence of m.

Surely this is something very simple, or it wouldn't be omitted in a textbook proof...

np = a^2 + b^2 + 1^2 + 0^2, so such an m <= n exists.
See also [URL="http://planetmath.org/encyclopedia/ProofOfLagrangesFourSquareTheorem.html"]the proof here[/URL], just below lemma 3.

[QUOTE=J.F.;157443]np = a^2 + b^2 + 1^2 + 0^2, so such an m <= n exists.
See also [URL="http://planetmath.org/encyclopedia/ProofOfLagrangesFourSquareTheorem.html"]the proof here[/URL].[/QUOTE]

:down: Yeah, that's it. Somehow I lost track of terms and was trying to make it a^2 + b^2 + 1^2 - np...

[QUOTE=CRGreathouse;157441]I started reading Nathanson's [i]Additive Number Theory[/i] and I hate to admit that I'm stuck on page 6. The proof is of Lagrange's Four-Square theorem (of course), and here's the setup toward showing that p can be represented as the sum of four squares.

It has just been proven that there is some integer n such that a^2 + b^2 + 1 = np, with 0 < n < p. The proof continues, "Let m be the least positive integer such that mp is the sum of four squares... we must show that m = 1."

How do we know that such a number exists? I mean, without assuming the four-square theorem itself. :razz:

And no, the proof doesn't say -- the RAA proof that m = 1 starts "Suppose not. Then 1 < m < p.", which also assumes the existence of m.

Surely this is something very simple, or it wouldn't be omitted in a textbook proof...[/QUOTE]

I prefer a much simpler proof (but more modern).

(1) Show that the quaternions are a UFD.
(2) Factor M \in Z over the quaternions; Then the representation of M
as the sum of 4 squares becomes immediate.

I've seen that proof, Dr. Silverman -- though not quite in that elegant form. That's related to the four-square identity of Euler, isn't it.

Does the proof generalize to other additive problems, though? I'm really interested in Romanov's problem on the sum of primes and powers of two (and variants therof).