Rating:

## Solution

The solution for this is straightforward by using __extended euclidean algorithm__.

Given `ct1` and `ct2`, we observe that

```

ct1*ct2 = (flag^13)*(flag^15) = flag^(13+15) mod n

```

And we can get any linear combination of 13 and 15

```

(ct1^a)*(ct2^b) = (flag^(13*a))*(flag^(15*b)) = flag^(a*13+b*15) mod n

```

And for negative values of `a` and `b` we get the inverse of ciphertext.

```

modinv(ct1, n)^k = (ct1^-1)^k = flag^(13*-k) mod n

```

With that, we simply need to find `a` and `b` such that

```

a*13 + b*15 = 1

```

And this has a solution because 13 and 15 are __coprime__. We use the __extended euclidean algorithm__ to find `a` and `b`.

__For implementation details see the URL__

Original writeup (https://github.com/pberba/ctf-solutions/blob/master/20181109_squarectf/crypto-flipping-bits/README.md).