## math contest question in hongkong

Posted: May 10, 2013 in Mathematics

Question:

Let $[x]$ be the largest integer not greater than $x$.  How many numbers of integer $n$ are there such that $\left[\dfrac{n^2}{3}\right]$ are prime number?

Since a set of integers can be expressed as the forms,

$3k, 3k+1, 3k+2, \forall k\in\mathbb{Z}$

Then we have 3 cases to consider,

Case 1, $n=3k$

$\left[\dfrac{(3k)^2}{3}\right] = \left[3k^2\right] = p$

Since p is prime,

$k=1, \therefore n=3$

Case 2, $n=3k+1$

$\left[\dfrac{(3k+1)^2}{3}\right] = \left[\dfrac{9k^2+6k+1}{3}\right] \\ = \left[3k^2+2k+\dfrac{1}{3}\right]=3k^2+2k=k(3k+2)= p$

Since p is prime,

$k=1, \therefore n=4$

Case 3, $n=3k+2$

$\left[\dfrac{(3k+2)^2}{3}\right] = \left[\dfrac{9k^2+12k+4}{3}\right] \\ = \left[3k^2+4k+\dfrac{4}{3}\right] = 3k^2+4k+1 = (3k+1)(k+1) = p$

Since p is prime, k has no solution.

Therefore, the only possible integers n are 3 or 4.