Fermat two square and Lagrange four square via Hardy-Littlewood circle method [closed]

Question

Fermat two square: An odd prime p is expressible as

${\displaystyle p=x^{2}+y^{2},\,}$

with $x, y$ integers, if and only if

${\displaystyle p\equiv 1{\pmod {4}}.}$

Lagrange four square: Every positive integer can be written as the sum of at most four squares.

Legendre's three-square: a natural number can be represented as the sum of three squares of integers

${\displaystyle n=x^{2}+y^{2}+z^{2}}$

if and only if $n$ is not of the form ${\displaystyle n=4^{a}(8b+7)}$ for integers a and b.

I know there are many proofs for these theorems, but my question is that how it is possible to prove these theorems by Hardy-Littlewood circle method. If not possible to prove it completely (for all numbers stated in theorems) how can we prove them for large enough numbers are possible? Is there any source that is already proved there?

Thanks.

Answer 1

Let me sketch a proof using Fourier analysis that I quite like, although perhaps not exactly what I would call the Hardy-Littlewood circle method.

Let $r(n)$ denote the number of representations of $n$ as the sum of two squares. Fermat's theorem is a corollary to the more general statement that $r(n)=4(d_1(n)-d_3(n))$, where $d_1(n)$ (resp. $d_3(n)$) denotes the number of divisors congruent to $1$ mod $4$ (resp. $3$ mod $4$).

If we consider the formal power series $$\theta(z) = \sum_{n=1}^{\infty} q^{zn^2}$$ and $$C(z) = 4\sum_{n=1}^{\infty}(d_3(n) q^{zn} - d_4(n) q^{zn}) = 2 \sum_{n=-\infty}^{\infty} \frac{1}{q^{zn}+q^{-zn}}$$ then the general claim reduces to the statement that $$\theta(z)^2 = C(z).$$ Now if we let $q^{zn} := e^{2 \pi i n z}$, then both of the functions above are easily seen to be holomorphic in the upper half plane.

Now notice that along the imaginary axis ($z=it$) $\theta(z) = \theta(it) = \sum_{n=1}^{\infty} e^{-\pi n^2 t}$, and this function can be obtained by applying Poisson summation to the Gaussian function. Similarly $$C(z) = \sum_{n=-\infty}^{\infty} \frac{1}{\cos(n\pi z)}.$$ Again along the imaginary axis ($z=it$) the imaginary axis $\cos(n\pi z)$ is $\cosh(n \pi t)$.

Now using the fact that $e^{-\pi x^2}$ and $1/\cosh(x)$ are both their own Fourier transform allows one to see, via Poisson summation, that these two functions satisfy certain functional equations. A bit of complex analysis using these functional equations then allows one to prove that $\theta(z)^2 = C(z)$ giving the claimed result.

One proof along these lines is given at the end of Stein and Shakarchi's complex analysis book.