For how many positive integers $n$$$\left \lfloor \frac{n^2}{3} \right \rfloor$$is a prime number?

$\left \lfloor \right \rfloor$ denotes the floor function.

Question

For how many positive integers $n$$$\left \lfloor  \frac{n^2}{3} \right \rfloor$$is a prime number?

$\left \lfloor  \right \rfloor$ denotes the floor function.

ganeshie8 · Answer

my initial guess is to use little fermat
for $(n,3)=1$ we have
$$n^2\equiv 1\pmod{3}$$

anonymous · Answer

what comes after that?

ganeshie8 · Answer

$$\left \lfloor  \frac{n^2}{3} \right \rfloor = \left \lfloor  \frac{3k+1}{3} \right \rfloor = k$$ ?

anonymous · Answer

well, that's right, but how do we know that $k$ is a prime or not?

anonymous · Answer

Use elementary methods. Just simplify the expression.

ganeshie8 · Answer

$$n^2 = 3k+1 \implies 3k = n^2-1=(n-1)(n+1)$$

p0sitr0n · Answer

For a general field $$\mathbb{F}=\mathbb{R}$$ this simplification won't work tho , since the floor function has an input interval that equals all the same value, i.e. $$\lfloor5.4\rfloor=\lfloor5.2\rfloor \nrightarrow 5.4 = 5.2$$, so careful

ganeshie8 · Answer

that yields $\large k\in \{3,5\}$ 

consequently $n\in \{4\}$

ganeshie8 · Answer

for $(n,3)\ne 1$, $\left \lfloor  \frac{n^2}{3} \right \rfloor =\left \lfloor  \frac{(3k)^2}{3} \right \rfloor  = 3k^2$ is always composite except for $k=1$

ganeshie8 · Answer

Overall $n\in\{3,4\}$

anonymous · Answer

how about $n=3k+2$

anonymous · Answer

and $n=3k+1$?

anonymous · Answer

sry for late response, I was out

ganeshie8 · Answer

little fermat covers both n=3k+1, 3k+2 right

anonymous · Answer

right :)

ganeshie8 · Answer

would love to see an alternate method as my method above is pretty hacky

ganeshie8 · Answer

another way to look at it is : $n^2$ can never be $3k+2 $ because $(3k\pm 1)^2 = 3M+1$. so $n^2\equiv 0,1\pmod{3}$