With some calculations we can get p and q

a2 + b2 = c2

(p + q)2 + (p - q)2 = c2

2p2 + 2q2 = c2

ab = 2A

(p + q) (p - q) = 2A

p2 - q2 = 2A

2p2 - 2q2 = 4A

2p2 + 2q2 = c2

4p2 = 4A + c2

And we got the p! Now we can get q easily with plugging p into one of the equation. So let’s compute them!

import gmpy2
c_square = 41027546415588921135190519817388916847442693284567375482282571314638757544938653824671437300971782426302443281077457253827782026089649732942648771306702020
A=1780602199528179468577178612383888301611753776788787799581979768613992169436352468580888042155360498830144442282937213247708372597613226926855391934953064

p = int(gmpy2.isqrt(c_square + 4 * A) // 2)
q = int(gmpy2.isqrt(p**2 - 2 * A))
print("p: {}".format(p))
print("q: {}".format(q))

Now let’s decrypt the message!

from Crypto.Util.number import inverse

n = p * q
e=0x10001
enc=825531027337680366509171870396193970230179478931882005355846498785843598000659828635030935743236266080589740863128695174980645084614454653557872620514117

phi = (p -1) * (q - 1)
d = inverse(e, phi)

m = pow(enc, d, n)
print((hex(m))[2:-1].decode('hex'))

Flag: castorsCTF{n0th1ng_l1k3_pr1m3_numb3r5_t0_w4rm_up_7h3_3ng1n3s}