Is the answer small?

Let $16p+1=m^3$ for some integer $m$ . Then, $16p=m^3-1=(m-1)(m^2+m+1)=(m-1)\big(m(m+1)+1\big)$ Since $m(m+1)$ is a product of two consecutive integers, it is even. Hence the second factor $\big(m(m+1)+1\big)$ is odd. Thus it must be that $16 | m-1$ . Let $m-1=16k$ , for some integer $k$ . Then, we have $p=k\big((16k+1)^2+(16k+1)+1\big)$ However, since $p$ is a prime, we must have $k=1$ . This implies $p=17^2+17+1=307$ It can be verified that $307$ is a prime and satisfies the required condition.

Write $16p+1=n^3$ or $16p=n^3-1$ $=(n-1)(n^2+n+1)$ . Now $n^2+n+1$ is odd for all $n$ . Consider the factorization of $n^2+n+1$ into (odd) primes: Since $16p$ has only one odd prime factor, $p$ , we must have $p=n^2+n+1$ and therefore $16=n-1$ . Thus $n=17$ and $p=307$ .

$16p + 1 = (n+1)^3 = n^3 +3n^2 + 3n + 1$ where n is a positive integer.

$16p = n^3+ 3n^2 + 3n = n(n^2 + 3n + 3)$

Is n odd or even? If n is odd, then $(n^2 + 3n + 3)$ is odd as well which means $16p$ is odd. This is absurd as p is an integer. So n has to be even. Even then, $(n^2+ 3n+ 3)$ is odd which tells us that n has to be divisible by 16. Let $n=16k$ where k is a positive integer.

$16p = 16k((16k)^2 + 3(16k)+ 3)$

$p = k(256k^2+48k+3)$ . For p to be prime, $k=1$ so it follows that $p = 256+48+3 = \boxed{307}$ which is indeed prime.

Did the same way.

16p = x^3 - 1 = (x-1)(x^2 + x + 1) = even times odd since x must be odd. 16p must only factor into 16 times p, since all other factors would be even times even or odd times even. So, x-1 = 16, resulting in x = 17. We get p = x^2 + x + 1 = 17^2 + 17 + 1 = 307.

Let 16 p+ 1 = x³

so x³ – 1= 16p

(x – 1)(x² + x + 1) = 16p

p = (x – 1)(x² + x + 1)/16

as p is integer (x – 1)(x² + x + 1) is divisible by 16

this possible when x = 17, 33, 49 ..,

all above expect 17 are composite. so p = 17

after solving , x = 17 , p = 307

16 p + 1 = m^3

So

p does not equal 2, i e p is odd

16 p = (m - 1)(m^2 + m + 1) ................ (1)

(m^2 + m + 1) = [m(m + 1) + 1] = odd number,

So

(m -1) is divisible by 16

i e

m - 1 = 16 k ...............................................(2)

Substituting from (2) in (1) we get

p = k (m^2 + m + 1) , but p is prime

So

k = 1

m = 17

p = 17^2 + 17 + 1 = 307

I wrote a program which checked whether a number was a prime number. If it was, it checked whether that number multiplied by 16 and added to 1 equaled to the cube of numbers from 0 to 50. def main(): for i in range(2,1000): if(isPrime(i)): for n in range(50): num = n n n comp = 16*i + 1 if(num == comp): print(i) def isPrime(n): if(n == 2 or n == 3) : return True for i in range(2,n): if(n % i == 0): return False return True

Also 2015 AIME I #3.