Does anyone know a proof of the 3 square theorem which uses Cassels' Lemma: n can be expressed in 3 squares of rationals iff n can be expressed in 3 squares of integers.



3 square theorem
Probably worth pointing out a certain distinction. The quoted property is Pete L. Clark's ADC property, see:


[Must a ring which admits a Euclidean quadratic form be Euclidean?](https://mathoverflow.net/questions/39510/must-a-ring-which-admits-a-euclidean-quadratic-form-be-euclidean) 


For positive forms in three variables with integer coefficients, this is stronger than L. E. Dickson's "regularity," which means that the form integrally represents any number represented by some form in the same genus. The ADC property is not implied by class number one here, nor does ADC guarantee class number one. I sent Pete the 103 positive ternary ADC forms; there are 794 positive ternaries alone in their genera. Also, thanks to Pete for actually checking something I foolishly assumed, there are several ADC forms not alone in their genera, such as $\langle 1,1,3,0,1,0 \rangle$ or $g(x,y,z) = x^2 + y^2 + 3 z^2 + z x,$ of discriminant 11. 
 I also gave an example for the 810 failures, each such regular form (including 14 not proved yet, after B-K Oh proved 8) represented some $p^2 n$ without representing $n,$ where the prime $p$ divides the Brandt-Intrau-Watson discriminant.


So the subtle point is that the ADC property in "Cassels's lemma" includes primes that divide the discriminant, which in the case of three squares is $-4,$ or taken as 4 by me and J. L. Lehman. The fact that the sum of three squares represents $n$ if and only if it represents $4n$ is, well, among regular forms, an artifact of a fortunate arrangement of local conditions, in this case the single "congruence obstruction" $4^k(8 m + 7).$ 


For example, the regular form $\langle 1, 1, 3, 1, 1, 1 \rangle$ or $g(x,y,z) = x^2 + y^2 + 3 z^2 + y z + z x + x y,$ integrally represents 8 but not 2. Oh, the discriminant of this form is 8, so the only prime available to distinguish between regularity and the stronger ADC property is 2.


Here is an easier one: $x^2 + y^2 + 4 z^2$ is regular, related in an evident way to the sum of three squares, and is not ADC as it integrally represents 12 but not 3. 


