I need some help with Uspensky's proof on the sum of three squares found in "Elementary Number Theory" by Uspensky and Heaslet. I have gotten to the end of the proof and do not understand a few statements found at the end. I was wondering if anyone was familiar with the proof. In it he defines $\displaystyle T(n) $ to be the total solutions to

$\displaystyle 4n+1=ab+(a+b-2)c$

$\displaystyle 4n+1=ab+(a+b+2)c$

With $\displaystyle a+b \equiv 0 \pmod 4$ and $\displaystyle a,b,c$ all odd.

Let $\displaystyle N_3(n)$ be the number of representations of $\displaystyle n$ as the sum of three squares.

He shows $\displaystyle 2[T(4d-1^2)+T(4d-3^2)+T(4d-5^2)+\ldots ]=N_3(4d-1^2)+N_3(4d-3^2)+N_3(4d-5^2)+\ldots$

Which I understand where that comes from. What I am confused on, is he claims that since this identity is true for all odd values of $\displaystyle m$ that we must have

$\displaystyle 2T(4d-1^2)=N_3(4d-1^2)$

Why must we have that? Any help would be greatly appreciated.