A Prime, Probably

It is given that $p$ is a prime number. Thus, $p$ is either $2$ or some odd number.

Checking the equation for $p=2$ , $32 \times 2 + 1 = 65$ . However, $4^3=64<65<125=5^3$ .

Hence, any $p$ satisfying the condition needs to be an odd number.

Let $p$ be the required solution, then

$32p+1=r^3 \implies 32p=r^3-1=(r-1)(r^2+r+1)$ . Performing a prime factor decomposition of the LHS gives, $2^5p=(r-1)(r^2+r+1)$ .

Note that $32p+1$ is always an odd number. Hence, $r^3$ is odd, and consequently, $r$ is also an odd number. This implies that $r-1$ is even and $(r^2+r+1)$ is odd.

Thus, $2^5p=(r-1)(r^2+r+1)$ would necessitate that $2^5=(r-1)$ and $p=r^2+r+1$ . The reason for this is trivial as the prime factor decomposition of $r^2+r+1$ cannot include $2$ .

$(r-1)=2^5 \implies r=33$ , which would mean $p=33^2+33+1=\boxed{1123}$ which is a prime number, and is the ONLY solution.

I think $1123$ is not only the least such prime but it is also the only one.