anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ganeshie8
  • ganeshie8
my initial guess is to use little fermat for \((n,3)=1\) we have \[n^2\equiv 1\pmod{3}\]
anonymous
  • anonymous
what comes after that?
ganeshie8
  • ganeshie8
\[\left \lfloor \frac{n^2}{3} \right \rfloor = \left \lfloor \frac{3k+1}{3} \right \rfloor = k\] ?

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anonymous
  • anonymous
well, that's right, but how do we know that \(k\) is a prime or not?
anonymous
  • anonymous
Use elementary methods. Just simplify the expression.
ganeshie8
  • ganeshie8
\[n^2 = 3k+1 \implies 3k = n^2-1=(n-1)(n+1)\]
P0sitr0n
  • P0sitr0n
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
  • ganeshie8
that yields \(\large k\in \{3,5\}\) consequently \(n\in \{4\}\)
ganeshie8
  • ganeshie8
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
  • ganeshie8
Overall \(n\in\{3,4\}\)
anonymous
  • anonymous
how about \(n=3k+2\)
anonymous
  • anonymous
and \(n=3k+1\)?
anonymous
  • anonymous
sry for late response, I was out
ganeshie8
  • ganeshie8
little fermat covers both n=3k+1, 3k+2 right
anonymous
  • anonymous
right :)
ganeshie8
  • ganeshie8
would love to see an alternate method as my method above is pretty hacky
ganeshie8
  • ganeshie8
another way to look at it is : \(n^2\) can never be \(3k+2 \) because \((3k\pm 1)^2 = 3M+1\). so \(n^2\equiv 0,1\pmod{3}\)

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