<s>[INST] Find all the natural numbers $n$ such that $\left\lfloor\frac{n^2}{3}\right\rfloor$ is prime How many natural numbers $n$ exist for which $\left\lfloor\frac{n^2}{3}\right\rfloor$ is prime?
I thought a lot but I don't  how to calculate the correct part.
 [/INST] So it's pretty clear that $n = 3k$ only works for $k=1$ i.e. $n=3$.
For $n = 3k+1$ we have that $n^2 = 9k^2 + 6k + 1$, so $\lfloor \frac{n^2}{3}\rfloor = 3k^2 + 2k$. If $k$ is even then this number is divisible by $2$. If $k$ is odd, then this number is divisible by $k$, which means that we still have a solution when $k=1$ i.e. $n=4$.
For $n=3k-1$ we can apply a similar technique and get $n^2 = 9k^2 - 6k + 1$ so $\lfloor \frac{n^2}{3}\rfloor = 3k^2 - 2k$. In this case $k=1$ gives us $1$ which is not a prime.
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