Approaches to Solving Quadratic Diophantine Equations of the Form $x^2 + y^2 = k$

Question

I am exploring the solutions of quadratic Diophantine equations in the following form:

$$ x_1^2 + y_1^2 = x_2^2 + y_2^2 = \dots = x_n^2 + y_n^2 = k, $$

where $x_1, \dots, x_n, y_1, \dots, y_n, k \in \mathbb{N}$. For the similar case of $x_1^2 - y_1^2 = x_2^2 - y_2^2 = \dots = x_n^2 - y_n^2 = k$, it's relatively straightforward to express each term as $(x_i - y_i)(x_i + y_i) = k$ and find pairs $(x_i, y_i)$ from the permutations of $k$'s prime multipliers, determining $n$ in the process. However, I am struggling to find a similar approach for the sum case.

I have searched for similar problems but haven't found much that deals with this specific form. Are there known methods or approaches to tackle this kind of problem? Any pointers towards relevant literature or examples would also be greatly appreciated.

Answer 1

There is an Identity of degree four & is given below:

$(4m^4-n^4)^2+(4m^2n^2)^2$=

$(8m^3n-4mn^3)^2+(4m^4-8m^2n^2+n^4)^2$=

=$(p^2+q^2)$=$(r^2+s^2)$=$(4m^4+n^4)^2$

where:

$p=(n^2+2mn)(2m^2-2mn+n^2)$

$q=(2m^2+2mn)(2m^2-2mn+n^2)$

$r=(n^2-2mn)(2m^2+2mn+n^2)$

$s=(2m^2-2mn)(2m^2+2mn+n^2)$

for, $(m,n)=(2,1)$ we get:

$65^2=63^2+16^2=56^2+33^2=25^2+60^2=39^2+52^2$

Answer 2

COMMENT.- $(1)$ First at all, the integer $k$ must be of the form $k=\prod p_i^{n_i}q_j^{2n_j}$ where the primes $p_i$ are conguent to $1$ modulo $4$ and the primes $q_j$ are conguent to $-1$ (or equivalently to $3$) modulo $4$. Otherwise $k$ cannot be a sum of two squares.

$(2)$ Next, the general solution of the equation $x^2+y^2=z^2+w^2$ is known to be given by the parameterization $$2x=rX+sY\\2y=rY-sX\\2z=rX-sY\\2w=rY+sX$$ where $r,s$ are arbitrary integers and $X,Y$ are two parameters (it is easy to prove this general solution because we have clearly an identity directly verifiable.

$(3)$ What remains depends on $n$: the larger $n$ is, the greater the difficulty will be in conditioning the step $(2)$ to your purpose.