What is $x$ in the formula? What is $x$ (assume it's integer) in $512p+ 1 = x^3$, where $p$ is a prime number. 
Attempt:
$$a^3 - b^3 = (a-b) (a^2 + ab+b^2) \Longrightarrow 512p = x^3 -1 = (x-1)(x^2+x+1).$$
Here I got stuck. Am I suppose to plug in $p$ and try it one by one?

What about $16p + 1 = x^3$??
 A: $512p=2^9p=x^3-1=(x-1)(x^2+x+1)$. Note that $x^3$ is odd (and hence $x$ is odd) since it's an even number plus $1$. When $x$ is odd, then $x^2+x+1$ is also odd and $x-1$ is the only even factor which implies that $p=x^2+x+1$ and $2^9=x-1$ (otherwise $p$ would not be prime or odd). Hence, the only possible solution could be $2^9=x-1\implies x=2^9+1\implies p=(2^9+1)^2+2^9+2$ which is not prime (it's divisible by $7$). Hence, there are no solutions where $p$ must be prime.
A: You are on the right track. Since $512p+ 1 = x^3$, we know that $x^3$ is odd, so that $x$ must be odd. 
When you have factored it into $512p = x^3 -1 = (x-1)(x^2+x+1)$, since $x$ is odd, $x^2+x+1$ is odd, so $x-1$ must be some multiple of 512. However, if $x-1$ is not exactly 512, $p$ won't be a prime. So we have $x-1 = 512$, hence $x=513$. Substitute this solution back, you will find that $p$ is not a prime, so this equation has no solution when $p$ is prime.
A: $512p + 1 = x^3$
$512p = x^3 -1$
$512p = (x-1)(x^2 + x + 1)$
$8^3p = (x-1)(x^2 + x + 1)$
If $x$ is even then $x^2 + x + 1$ is odd, and $x-1$ is odd and LHS is odd.  That's impossible.
So $x$ is odd.  So $x^2 + x + 1$ is odd.  So $512|x-1$
$p = \frac {x-1}{512}(x^2 + x + 1)$.
If $x \le 0$ then $ \frac {x-1}{512}(x^2 + x + 1) < 0$ so that's a contradiction, so $x^2 + x + 1 > 1$
$p$ is prime. So it only has $1$ and $p$ as factors.  So $x^2 + x + 1 = p$ or $x^2 + x+1=1$.  The later is impossible so $\frac {x-1}{512} = 1$ and $x^2 + x+1 = p$... That is, if there is any answer at all.
So $x = 513$ and $p =513^2 + 513 +1$... if, such a number is prime.  If not, there is no solution.
A: suppose first: $x=2k,  k\in \mathbb{Z}$; then 
$$(2k-1)[2(k^2+k)+1] = 512p \\ \implies \text{No such } p \text{ exists where } (2k-1)  \& [2(k^2+k)+1] \text{ are odd numbers.}$$
suppose second: $x=2k+1, k\in \mathbb{Z}$; then 
$$(2k)[2(k^2+k)+1] = 512p $$
$$ (k)[2(k^2+k)+1] = 256p $$
$$ \text{where } [2(k^2+k)+1] \text{ is an odd number.} \implies k=256, p = 2(k^2+k)+1  \\ \implies p=263683 \quad \textbf{dividable by 7, so, not acceptable.} $$
$$\implies \text{No such } p \text{ exists.}$$
For: $16p+1 = x^3$
$$(x-1)(x^2+x+1) = 16p$$ 
where $x^2+x+1$ is an odd number, so: $x=17$ and $p=307$ which is a prime number.
