# Really Secure Algorithm
## Description
I heard about RSA, so I took a go at implementing it.
[Secure.txt](Secure.txt)
## Solution
The n is actually a sqaure number.
```***factors found***
P386 = 16225510719965861964299051658340559066224635411075742500953901749924501886090804067406052688894869028683583501052917637552385089084807531319036985272636554557876754514524927502408114799014949174520357440885167280739363628642463479075654764698947461583766215118582826142179234382923872619079721726020446020581078274482268162477580369246821166693123724514271177264591824616458410293414647P386 = 16225510719965861964299051658340559066224635411075742500953901749924501886090804067406052688894869028683583501052917637552385089084807531319036985272636554557876754514524927502408114799014949174520357440885167280739363628642463479075654764698947461583766215118582826142179234382923872619079721726020446020581078274482268162477580369246821166693123724514271177264591824616458410293414647```
If n = p * p, phi(n) = p * (p-1)
Perform mod inverse on e and decrypt the message.
```d=modinv(e,(p-1)*p)import codecsprint(codecs.decode(hex(pow(c,d,n))[2:],'hex'))```
The flag is
```hsctf{square_number_time}```