Find the primes $p$ such that the equation: $x^{2} + 7x + 39 = 0 $ has a solution modulo $ p $.

Question

I have this question:

if $ p \neq 2 $ so the equation has a solution $ \iff 7^{2} -4\cdot 1\cdot 39 = -107 $ is a square modulo p .

now $ -107 $ is a square modulo p $ \iff (\frac{-107}{p}) = 1$ (legendre symbol) $

Now I got stuck because I can't find a way to decmopose 107 or any other way to continue from here. I really appreciate any help!!

Very similar to your previous question asked 4 minutes earlier? Only the table will be longer. Too long for this to come from a teacher IMO. — Jyrki Lahtonen, Aug 03 '16 at 21:58
It is indeed unpleasant, for it looks as if we need to know the quadratic residues of $107$. — André Nicolas, Aug 03 '16 at 22:08
@JyrkiLahtonen So what should I do now actually? what is the table you're talking about? — CnR, Aug 03 '16 at 22:14

score 1 · Answer 1 · answered Nov 29 '16 at 20:41

You can't do very much else than (find a) list (of) all quadratic residues modulo 107. Why, you ask? Well, with the help of multiplicity and quadratic residue theorem, the argument goes as follows:

$$(\frac{-107}{p})=(\frac{-1}{p})(\frac{107}{p})=(-1)^{\frac{p-1}{2}}(\frac{107}{p})$$ and $$(\frac{107}{p})(\frac{p}{107})=(-1)^{\frac{p-1}{2}\cdot\frac{107-1}{2}}=(-1)^{{\frac{p-1}{2}\cdot 53}}=(-1)^{\frac{p-1}{2}}$$ thus, assuming $(\frac{-107}{p})=1$, $$(\frac{p}{107})=(-1)^{\frac{p-1}{2}}(\frac{107}{p})=(-1)^{\frac{p-1}{2}}(-1)^{\frac{p-1}{2}}(\frac{-107}{p})=1$$ Hence, p is a quadratic residue modulo 107. This will give you 53 possible values x with $p\equiv x \mod 107$ as solutions.

Find the primes $p$ such that the equation: $x^{2} + 7x + 39 = 0 $ has a solution modulo $ p $.

1 Answers1