Suppose that I am given an odd prime $p$. In addition, suppose that $$\left(\dfrac{75}{p}\right) = -1, \left(\dfrac{93639}{p}\right) = 1.$$
I am solving for $\left(\dfrac{4179}{p}\right).$
I have utilized the Legendre symbol, so this does not mean division. I was trying to see if there were any relations between $4179,75,$ and $93639$, such as multiplying $4179$ and $75$, but this did not work. so then, I began to find the prime factorization of $4179$ and got
$$4179 = 3 \cdot 7 \cdot 199$$
So we have $$\left(\dfrac{4179}{p}\right) = \left(\dfrac{3}{p}\right) \left(\dfrac{7}{p}\right) \left(\dfrac{199}{p}\right).$$
Now,
$$\left(\dfrac{75}{p}\right) = \left(\dfrac{5}{p}\right) \left(\dfrac{5}{p}\right) \left(\dfrac{3}{p}\right) = \left(\dfrac{3}{p}\right) = -1.$$
So we have
$$\left(\dfrac{4179}{p}\right) = (-1) \left(\dfrac{7}{p}\right)\left(\dfrac{199}{p}\right)$$
Where can I go from here?
Major edit
It turns out that there was indeed a typo in the problem after all and my answer below will discuss how to carry out this problem correctly.
It turns out that there was a typo in the question, which caused some of us to be frustrated with the calculations, including myself. Here, instead of $4179$, it is $4719$ and we will see why this change in numbers is significant to our final result. We begin by factoring $4719$ into primes to yield
$$4719 = 3 \cdot 11^2 \cdot 13$$
Thus, we have
$$\left(\dfrac{4719}{p}\right) = \left(\dfrac{3}{p}\right) \left(\dfrac{11}{p}\right) \left(\dfrac{11}{p}\right) \left(\dfrac{13}{p}\right) = \left(\dfrac{3}{p}\right) \left(\dfrac{13}{p}\right)$$
Next, we do the prime factorization of $75$ and have $$75 = 3 \cdot 5^2$$
Thus, we have
$$\left(\dfrac{75}{p}\right) = \left(\dfrac{3}{p}\right) \left(\dfrac{5}{p}\right) \left(\dfrac{5}{p}\right) =\left(\dfrac{3}{p}\right) = -1$$ since $\left(\dfrac{75}{p}\right) = -1$ was by hypothesis. Thus, we have
$$\left(\dfrac{4719}{p}\right) = (-1) \left(\dfrac{13}{p}\right)$$
Now, we find the prime factorization of $93639$ to yield $$93639 = 3 \cdot 7^4 \cdot 13$$.
Hence, we have
$$\left(\dfrac{93639}{p}\right) = \left(\dfrac{3}{p}\right) \left(\dfrac{7}{p}\right) \left(\dfrac{7}{p}\right) \left(\dfrac{7}{p}\right) \left(\dfrac{7}{p}\right) \left(\dfrac{13}{p}\right) = \left(\dfrac{3}{p}\right) \left(\dfrac{13}{p}\right)$$ but $\left(\dfrac{3}{p}\right) = -1$ and $\left(\dfrac{93639}{p}\right) = 1$. Thus, we have
$$1 = -1 \left(\dfrac{13}{p}\right) $$
Thus, $\left(\dfrac{13}{p}\right) = -1$. Thus, we have
$$\left(\dfrac{4719}{p}\right) = (-1)(-1) = 1.$$
QED