2025/019)Find prime p such that 16p+1 is a perfect cube

$16p+1$ is perfect cube so let is be $x^3$

So $x^3-1 = 16p$

Or $(x-1)(x^2+x+1) = 16p$

Now $x^2+x+1$ is is odd so $16| x-1$

So $(x-1) = 16m$ and $x^2+x+1 = n$ where m and n are odd

So p = mn

As p is odd $m = 1, n = p$  or $m = p, n= 1$

n= 1 gives $x^2+x+1 = 1 $ or x = 0 which is not possible

so m= 1 and x= 17 giving n = 307 wihich is a prime

so $p = 307$