This is the first cryptography challenge. The code itself is not very lengthy, but its functionality is quite "uncommon" as it utilizes an external function to generate four integers a, b, c, and d, which have a relationship with n: `a^2 + b^2 = n`, `c^2 + d^2 = n````pywhile True:	try:		key = RSA.generate(2048)		a,b,c,d = magic(key)		break	except:		passassert a**2 + b**2 == key.nassert c**2 + d**2 == key.n```At this point, by using the `Brahmagupta–Fibonacci` method, you can solve the equation following these steps:
$$\begin{align}& a^2 + b^2 = c^2 + d^2 = n \newline& (a^2 + b^2)(c^2 + d^2) = n^2 = (pq)^2 \newline& (ac + bd)^2 + (ad - bc)^2 = p^2 q^2 \newline& q^2 = s^2 + t^2 \newline& (ac + bd)^2 + (ad - bc)^2 = (p \cdot s)^2 + (p \cdot t)^2 \newline& ps = a \cdot c + b \cdot d \newline& pt = a \cdot d - b \cdot c \newline& p = \text{gcd}(ps, pt) \newline& q = \frac{n}{p} \newline\end{align}$$