Determine the integers $a$ such that the congruence $x^4 \equiv a \pmod p$ has a solution for $p = 7, 11, 13$ Determine the integers $a$ such that the congruence $x^4 \equiv a \pmod p$ 
has a solution for $p = 7, 11, 13$

This looks similar to previous problem but kinda tricky. I'm not sure where to start... appreciate any help thanks!
 A: My recommendation: in each case, find a generator of the multiplicative group of the field. For instance for $p=7$, you can use $g=3$, and then you get $g^0=1$, $g^1=3$, $g^2=2$, $g^3=6$, $g^4=4$, $g^5=5$, $g^6=1$. With this “logarithm table”, you can answer all possible multiplicative questions of this sort.
A: First $x^4 \equiv 0 \pmod{p}$ if and only if $x \equiv 0 \pmod{p}$.
Next, you can prove that for invertible elements $x^4 \equiv y^4 \pmod{p}$ if and only if $(xy^{-1})^4 \equiv 1 \pmod{p}$.
Using the existence of primitive roots, you can prove that for odd primes the equation
$$z^4 =1 \pmod{p}$$
has 4 solutions when $4| p-1$ and $2$ solutions otherwise.
Therefore 
$$f: U(\mathbb Z/p \mathbb Z) \to U(\mathbb Z/p \mathbb Z) \\
f(x)=x^4$$
takes $\frac{p-1}{4}$ values when $4|p-1$ and $\frac{p-1}{2}$ values when $4 \nmid p-1$.
Adding back $0$ we get that
Therefore 
$$f: \mathbb Z/p \mathbb Z \to\mathbb Z/p \mathbb Z \\
f(x)=x^4$$
takes $\frac{p-1}{4}+1$ values when $4|p-1$ and $\frac{p-1}{2}+1$ values when $4 \nmid p-1$.
Therefore


*

*for $p=7$ there are thus $4$ values in the range of $f$.

*for $p=11$ there are thus $6$ values in the range of $f$.

*for $p=13$ there are thus $4$ values in the range of $f$.
And for each $p$ it is very easy to find those values. Just plug in numbers in $f$ until you get as many values as the range (and don't forget that 0,1 are in the range so you only need to find 2,4 respectively 2 more values). 
The problem is asking you to find all the $a$'s which are in all three ranges, and the Chinese Remainder Theorem tells you how to find all of them. It also tells you that there are $4 \times 6 \times 4$ distinct values $\pmod{7 \cdot 11 \cdot 13}$ (YIKES).
P.S. For the solution you can simply ignore the first part, and simply calculate the range of $f(x)=x^4$ for each $p$. The first part of the problem only calculates how many elements are in the range, so it can save some time when you start plugging numbers. 
