EVEN RSA CAN BE BROKEN???¶
On requesting multiple times, we notice that N is always even
p = 2
q = N//2
phi = (p-1)*(q-1)
d = inverse(e, phi)
m = pow(c, d, N)
print(long_to_bytes(m).decode('utf-8'))
On requesting multiple times, we notice that N is always even
p = 2
q = N//2
phi = (p-1)*(q-1)
d = inverse(e, phi)
m = pow(c, d, N)
print(long_to_bytes(m).decode('utf-8'))