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 $$\left(\dfrac{4179}{p}\right) = \left(\dfrac{3}{p}\right) \left(\dfrac{7}{p}\right) \left(\dfrac{199}{p}\right).$$


$$\left(\dfrac{75}{p}\right) = \left(\dfrac{5}{p}\right) \left(\dfrac{5}{p}\right) \left(\dfrac{3}{p}\right) = \left(\dfrac{3}{p}\right) = -1.$$

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.

  • $\begingroup$ Is there any way of using the fact that $93639=3\cdot7^4\cdot13$ ? $\endgroup$ – Lubin Apr 10 '18 at 3:31
  • $\begingroup$ I have tried doing just that. In other words, $\left(\dfrac{93639}{p}\right) = \left(\dfrac{3}{p}\right) \left(\dfrac{7}{p}\right) \left(\dfrac{7}{p}\right) \left(\dfrac{7}{p}\right) \left(\dfrac{7}{p}\right) \left(\dfrac{13}{p}\right) = \left(\dfrac{3}{p}\right) \left(\dfrac{13}{p}\right).$ $\endgroup$ – John Bradshaw Apr 10 '18 at 3:34
  • $\begingroup$ Thus $(\frac{13}p)=-1$, and since $13\equiv1\pmod4$, we have $(\frac p{13})=-1$, which means that $p\equiv2,5,6,7,8,\text{ or }11\pmod{13}$, though I don’t see how this helps. $\endgroup$ – Lubin Apr 10 '18 at 3:40
  • $\begingroup$ This is where I had issues and turned to MSE. I double-checked the numbers I originally had when I asked my question and they are correct. $\endgroup$ – John Bradshaw Apr 10 '18 at 3:45
  • 3
    $\begingroup$ You simply don't have enough information to solve this. $\endgroup$ – Angina Seng Apr 10 '18 at 3:48

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

$$\left(\dfrac{4719}{p}\right) = (-1)(-1) = 1.$$


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