2022/076) Find prime p such that $16p+1$ is a cube

We have $16p + 1$ is a cube say $x^3$

So $16p= x^3-1 = (x-1)(x^2+ x + 1)$

As $x^2+x+1 = x(x+1) +1$ is odd we must have

$16 | x-1$

So $x = 16k+1$ for some integer k

so $p = k(x^2 + x + 1)$

the above cannot be prime unless k =1 and this gives x = 17 and p = 307

We need to see if  307 is prime and it is so

Hence the only solution is p = 307 (giving $16 * 307 +1  = 17^3$