$A$ is a sum of two postive integer squares? if $x,y,z,w$ be postive integer,and such $x^2+y^2$ is prime number,and  $A=\dfrac{w^2+z^2}{x^2+y^2}\in N^{+}$
show that
$A$ is a sum of  two  postive integer squares?
maybe 
$$A=\dfrac{(w^2+z^2)(x^2+y^2)}{(x^2+y^2)}=\dfrac{(wx+zy)^2}{(x^2+y^2)^2}+\dfrac{(wy-xz)^2}{(x^2+y^2)^2}$$
But I can't prove $\dfrac{wx+zy}{x^2+y^2},\dfrac{(wy-xz)}{x^2+y^2}\in Z$
 A: Approach 1
In this answer, it is shown that a number, $n$, can be written as the sum of two squares if and only if each prime in the prime factorization of $n$ which is $\equiv3\pmod4$ appears with even exponent. Thus, if
$$
\frac{w^2+z^2}{x^2+y^2}\in\mathbb{Z}
$$
each prime $\equiv3\pmod4$ in $w^2+z^2$ appears with even exponent, and each prime $\equiv3\pmod4$ in $x^2+y^2$ appears with even exponent. Therefore, each prime $\equiv3\pmod4$ in the quotient must appear with even exponent, and therefore, the quotient can be written as a sum of two squares.

Approach 2
Here is an alternate approach using the fact that $x^2+y^2$ is assumed to be a prime.
Since $x^2+y^2$ is a prime number, $x+iy$ and $x-iy$ are Gaussian primes. Thus, because
$$
x+iy\mid(w+iz)(w-iz)
$$
we can choose the sign of $z$ so that $x+iy\mid w+iz$, and then $x-iy\mid w-iz$, also.
Then
$$
\frac{w+iz}{x+iy}=a+ib\qquad\text{and}\qquad\frac{w-iz}{x-iy}=a-ib
$$
Thus,
$$
\frac{w^2+z^2}{x^2+y^2}=\frac{w+iz}{x+iy}\frac{w-iz}{x-iy}=(a+ib)(a-ib)=a^2+b^2
$$

Confirmation of the Hypothesis Regarding the Squares
Note in the last approach, we have that
$$
a+ib=\frac{w+iz}{x+iy}=\frac{(xw+yz)+i(xz-yw)}{x^2+y^2}
$$
and therefore,
$$
\frac{w^2+z^2}{x^2+y^2}
=\left(\frac{xw+yz}{x^2+y^2}\right)^2+\left(\frac{xz-yw}{x^2+y^2}\right)^2
$$
where
$$
\frac{xw+yz}{x^2+y^2},\frac{xz-yw}{x^2+y^2}\in\mathbb{Z}
$$
