# Prime Numbers

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.

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Hurkyl
Staff Emeritus
Gold Member
Try factoring.

After I factored the right side, i got $$x^2 = (v + 1)(v^2 - v + 1)$$ I dont see what to do next

Hurkyl
Staff Emeritus
Gold Member
You've found 2 factors, in terms of v.
Can you find all factors, in terms of x?

I dont understand the "all factors of x" part
you mean $$(x - 1)(x + 1) = v^3$$ ?

The only x that seem to work is x = 2

Hurkyl
Staff Emeritus
Gold Member
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...

Hurkyl said:
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 $$2 = v^2 - v + 1$$?

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Hurkyl
Staff Emeritus