Can Prime Numbers x Satisfy x^2 = v^3 + 1?

[Pandaren]

Hey everyone, I need help on this problem:

Find all prime numbers $$x$$ such that $$x^2 = v^3 + 1$$ for some integer $$v$$.

Thanks a lot for your help, i appreciated.

 

[Hurkyl]

Try factoring.

 

[Pandaren]

After I factored the right side, i got $$x^2 = (v + 1)(v^2 - v + 1)$$ I don't see what to do next

Thank you for your help

 

[Hurkyl]

You've found 2 factors, in terms of v.
Can you find all factors, in terms of x?

 

[Pandaren]

I don't understand the "all factors of x" part
you mean $$(x - 1)(x + 1) = v^3$$ ?

 

[Pandaren]

The only x that seem to work is x = 2

 

[Hurkyl]

Well, x^2 is the square of the prime, x, right? You know the prime factorization of x^2, so you can write down all of its factors.


Since v+1 is a factor, v+1 must be equal to one of the numbers on that list...

 

[Pandaren]

Well, x^2 is the square of the prime, x, right? You know the prime factorization of x^2, so you can write down all of its factors.


Since v+1 is a factor, v+1 must be equal to one of the numbers on that list...

you mean if 2 is a prime, then $$2 = x$$, $$2^2 = x^2$$
Then $$2^2 = (v + 1)(v^2 - v + 1)$$ Or $$2 * 2 = (v + 1)(v^2 - v + 1)$$
Therefore $$2 = v + 1$$ and $$2 = v^2 - v + 1$$?

 

[Hurkyl]

You know that (v+1)(v^2-v+1) is a factorization of x^2 into two factors, and you know that x * x is a factorization of x^2 into two factors, but you don't know that they're the same factorization.

However, you can write down all of the ways to factor x^2 into two factors, and you know that (v+1)(v^2-v+1) is going to be one of them.

 

[Pandaren]

Ohh I understand now thanks for your help