How to solve congruence $x^y = a \pmod p$? I'm having trouble solving this congruence:
$$x^{114} \equiv 13 \pmod {29}.$$
I thought that it made sense to try to solve it using this idea: "Suppose you want to solve the congruence $ x^y \equiv a \pmod p$ (we will assume for the moment
that $p$ is prime). Raise both sides to the power $z$ to obtain $x^{yz} \equiv a^z
\pmod p$. Now if we can
ﬁnd a $z $ such that $yz \equiv 1 \pmod {p − 1}$ then the solution of the congruence will be $x \equiv a^z \pmod p$."
So I set $114 z \equiv 1 \pmod {28}$. However, $\gcd(114, 28) = 2$ and I can't solve for the inverse using the Euclidean algorithm. Does that statement that I quoted even come in handy anywhere?
Next I simplified $x^{114}$ to $x^2$ by Fermat's theorem.
I know that $x^2 \equiv 13 \pmod {29}$ has a solution because of the Legendre symbol:
$$\left(\frac{13}{29}\right) \equiv 13^{14} \pmod {29} = 1$$
The only way I have learned to solve for square roots is when $p \equiv 3 \pmod 4$. Since this isn't the case, I'm at a loss as to how to find the square root. Any tips or hints?
 A: Here is my intuitive solution:
$$x^{114} \equiv 13 \pmod {29}$$
As shown above by the Fermat Little Theorem, the congruence is simplified to
$$x^{2} \equiv 13 \pmod {29}$$
Now we subtract $29$ from the right hand side of the congruence
$$x^{2} \equiv -16 \pmod {29}$$
We can notice that $-16=4*(-4)$, so we can say
$$4 \equiv 4 \pmod {29}$$
$$29-4 \equiv 25 \equiv -4 \pmod {29}$$
Now we multiply the last two congrunces
$$100 \equiv -16 \equiv 13 \pmod {29}$$
100 is a perfect square, thus the solution is $x=10$
A: Firstly, reduce the exponent immediately $\mod 28$ using Fermat's Theorem. So now you have to solve $x^2 = 13 \mod 29$ and since $13^{14} = 1 \mod 29$ you know there is a solution - actually there are two, namely $x$ and $29 - x$.
To find a solution in this simple case, just do a complete search up to $x = 14$. The result is $x = 10$. Anything else is a waste of time.
Now if you had to find $x$ such that
$$
x^{3000000000000000000350} = 345679012320987652917 \mod 1000000000000000000117
$$
that would be a different matter. That's also a problem of the form "solve $x^2 = a \mod p$ and $p = 1 \mod 4$." In that case use the algorithm proposed by Daniel Fisher. It's built into Mathematica.   
