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?

Answer:

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.

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