AIME Problem 7

Let $16p+1=a^3\\ \rightarrow 16p=a^3-1\\ \rightarrow 16p=(a-1)(a^2+a+1)$

Since $a^2+a+1=a(a+1)+1$ would always be odd, thus we know that 16 must divide $a - 1$ . If $a - 1 = 16 p$ , then we have $a ^2 +a + 1 = 1$ which gives us $a = 0$ , which we reject. Thus, $a - 1 = 16$ and $a^2 + a + 1 = p$ .

$a-1=16\\ \rightarrow a=17$

Now that we have $a=17$ , we get

$a^2 + a + 1 = p = 289+17+1 = \boxed{307}$

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By hypotesys $16p + 1 = (a+1)^3$ so

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

Now $a^2 + 3a + 3$ is odd. In fact $a^2 + 3a$ is even because $a^2 + 3a = a(a+3)$ . If $a$ is even $a+3$ is odd ad viceversa. So the product is between an even and an odd. So it is even. This means $2$ doesn't divide $a^2 + 3a + 3$ , and $16$ must divide $a$ .

We found that $a = 16b$ and substituting into our equation we have

$p = (256b^2+48b+3)b$

Since the natural $b$ divides a prime $p$ , it must be $p$ or $1$ . If it is $p$ then we have $p = (256p^2+48p+3)p > p$ . So it must be $b = 1$ and $p = 307$ .

Notice that 16p + 1 = a^3 , then 16p = a^3 - 1 = (a-1)(a^2+a+1)

Notice that a^2 + a + 1 = a(a+1) + 1 is always an odd number , so a^2 + a + 1 cannot be equal to 16.

Thus a - 1 = 16 , we get a = 17 , and p = a^2 + a + 1 = 307 , which is a prime

Since $16p+1$ is odd.

Let $16p+1=(2k+1)^3= 8k^3+12k^2+6k+1$

$\implies 8p=k(4k^2+6k+3)$

Since $p$ is prime and cannot be even (because when $p=2$ we note that $16 \times 2 + 1 =33$ is not a cube) and $4k^2+6k+3$ is odd

So $k$ must be equal to $8$ and

$p=4 \cdot 8^2 + 6 \cdot 8 + 3= \boxed{307}$

Also we check that $16 \times 307 + 1=4913 = 17^3$ is a perfect cube.

$x^3=16p+1\Rightarrow x^3 \equiv 1\pmod{16} \Rightarrow x \equiv 1\pmod{16}$ Trying the first few:

307 is a prime.

Since 16p+1 is an odd number, the integer it is the cube of, must be of the form 2n+1.

Solving 16p+1 = (2n+1)^3, we get 8p = n(4n^2 + 6n + 3)

The factor (4n^2 + 6n + 3) doesn't have integral factors. So, it can't be factorised further and, therefore, must be prime.

So, n must be equal to 8 and p must be equal to (4n^2 + 6n + 3).

Substituting n = 8 in p, we get p = 307.