Really Secure Algorithm

Given n, c, and e are big. Thrown n into factordb.com, it gave the same q and p, which means n=pow(p,2). As p and q are the same, phi=(p-1)*(q-1)=pow(p-1,2)

from Cryptodome.Util import number

n = <very big>
c = <very big>
e = <very big>

p = <result from factordb.com>

phi = pow(p-1,2)
d = number.inverse(e, phi)
ans = pow(c, d, p)
print(number.long_to_bytes(ans))

Decrypted message is the flag.

b'DawgCTF{sm@ll_d_b1g_dr3am5}'
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