Greatest integer mess

Firstly we have $n^2=3P$ or $n^2=3P+1$ or $n^2=3P+2$ .

• If $n=3k$ where $k$ is an integer, $n^2=9k^2=3(3k^2)$ , $P=3k^2$ .
• If $n=3k+1$ where $k$ is an integer, $n^2=9k^2+6k+1=3k(3k+2)+1$ , $P=k(3k+2)$ .
• If $n=3k-1$ where $k$ is an integer, $n^2=9k^2-6x+1=3k(3k-2)+1$ , $P=k(3k-2)$ .

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• First case: $3k^2$ can only be a prime when $k=1$ , which makes $P=3$ .
• Second case: $k(3k+2)$ can only be a prime when $k=1$ or $3k+2=1$ , and the latter is impossible since k would not be an integer, which makes $k=1$ the only possible case, and $P=5$ .
• Third case: $k(3k-2)$ can only be a prime when $k=1$ or $3k-2=1$ , and the two conditions are actually the same, which makes $P=1$ .

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Therefore, concluding from the three cases, the largest prime P is $\fbox5$ .