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anonymous
 one year ago
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.
anonymous
 one year ago
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.

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4my initial guess is to use little fermat for \((n,3)=1\) we have \[n^2\equiv 1\pmod{3}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what comes after that?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[\left \lfloor \frac{n^2}{3} \right \rfloor = \left \lfloor \frac{3k+1}{3} \right \rfloor = k\] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, that's right, but how do we know that \(k\) is a prime or not?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Use elementary methods. Just simplify the expression.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4\[n^2 = 3k+1 \implies 3k = n^21=(n1)(n+1)\]

P0sitr0n
 one year ago
Best ResponseYou've already chosen the best response.0For 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
 one year ago
Best ResponseYou've already chosen the best response.4that yields \(\large k\in \{3,5\}\) consequently \(n\in \{4\}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4for \((n,3)\ne 1\), \(\left \lfloor \frac{n^2}{3} \right \rfloor =\left \lfloor \frac{(3k)^2}{3} \right \rfloor = 3k^2\) is always composite except for \(k=1\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Overall \(n\in\{3,4\}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how about \(n=3k+2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sry for late response, I was out

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4little fermat covers both n=3k+1, 3k+2 right

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4would love to see an alternate method as my method above is pretty hacky

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4another way to look at it is : \(n^2\) can never be \(3k+2 \) because \((3k\pm 1)^2 = 3M+1\). so \(n^2\equiv 0,1\pmod{3}\)
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