In Quadratic Residues we learnt what it means to take the square root modulo an integer. We also saw that taking a root isn't always possible.

In the previous case when $p = 29$, even the simplest method of calculating the square root was fast enough, but as $p$ gets larger, this method becomes wildly unreasonable.

Lucky for us, we have a way to check whether an integer is a quadratic residue with a single calculation thanks to Legendre. In the following, we will assume we are working modulo a prime $p$.

Before looking at Legendre's symbol, let's take a brief detour to see an interesting property of quadratic (non-)residues.

Quadratic Residue * Quadratic Residue = Quadratic Residue
Quadratic Residue * Quadratic Non-residue = Quadratic Non-residue
Quadratic Non-residue * Quadratic Non-residue = Quadratic Residue

Want an easy way to remember this? Replace "Quadratic Residue" with $+1$ and "Quadratic Non-residue" with $-1$, all three results are the same!

So what's the trick? The Legendre Symbol gives an efficient way to determine whether an integer is a quadratic residue modulo an odd prime $p$.

Legendre's Symbol: $(a / p) \equiv a^{(p-1)/2} \mod p$ obeys:

$(a / p) = 1$ if $a$ is a quadratic residue and $a \not\equiv 0 \mod p$
$(a / p) = -1$ if $a$ is a quadratic non-residue $\mod p$
$(a / p) = 0$ if $a \equiv 0 \mod p$

Which means given any integer $a$, calculating $a^{(p-1)/2} \mod p$ is enough to determine if $a$ is a quadratic residue.

Now for the flag. Given the following 1024 bit prime and 10 integers, find the quadratic residue and then calculate its square root; the square root is your flag. Of the two possible roots, submit the larger one as your answer.

So Legendre's symbol tells us which integer is a quadratic residue, but how do we find the square root?! The prime supplied obeys $p = 3 \mod 4$, which allows us easily compute the square root. The answer is online, but you can figure it out yourself if you think about Fermat's little theorem.

Challenge files:
  - output.txt