For how many $n \in \mathbb{N}$ is $\sqrt{n^2+2379}$ natural?

Question

Here's my attempt at a solution: the expression $\sqrt{n^2+2379}$ is natural iff $$n^2 + 2379 = x^2, \quad \mbox{ for some } x \in \mathbb{N}.$$

Therefore $$(x+n)(x-n)=2379=3 \cdot 13 \cdot 61.$$

I try to represent $2379$ as a product of two natural numbers, this can be done in four ways:

$$2379=(3 \cdot 13 )\cdot 61$$ $$2379=3 \cdot (13 \cdot 61)$$ $$2379=(3 \cdot 61)\cdot 13$$ $$2379=1 \cdot 2379 $$

Comparing these options to $(x-n)(x+n)=2379$ produces four pairs, since $$(x-n)(x+n) = ab \implies x=(a+b)/2, \ n= (a-b)/2, \quad \mbox{ assuming } a>b.$$

The four $n$'s are then $11,85,395,1189$. Is my solution correct? Is there a way of doing this more theoretically, without knowing explicitly the prime factorization of $2379$?

Answer 1

Your solution is correct (I used Wolfram Alpha to double check the arithmetic). If one has a list of all solutions, it is easy to trace them back to the prime factors of $2379$, so essentially it is necessary to have the factorization in order to know all the solutions.

While the problem only asks for the number of solutions, this is equivalent to knowing the number of divisors of $2379$. I think it is widely believed that counting divisors is just as hard as factoring, but this has not been proven to my knowledge.