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Floor function and primes (Posted on 2006-12-09) Difficulty: 3 of 5
Determine the integer(s) n for which [n²/3] is a prime.
Note: [x] is the greatest integer ≤ x (floor function).

No Solution Yet Submitted by atheron    
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re: A complex proposal Comment 5 of 5 |
(In reply to A complex proposal by JLo)

Assuming you meant :

A complex integer is a prime if all of its factorizations into two integers contains exactly one of -1, 1, i or -i as a factor.

I belive that the complex primes are then the union of:
{The real primes congruent to 3 mod 4}*{1,i,-1,-i}
{a+bi where a^2+b^2 is a prime}

Then it appears that there is no nice solution like there is for the original problem.

4 and -4 are no longer solutions (5 = (2+i)(2-i))
3i and -3i are, but things get wierd when we look at  number of the form n=(ai-a) giving [n^2/3]=-i*ceiling((2/3)a^2)
There appear to be an unbounded number of such n that lead to primes, here are the ones I founf before stopping:

2i-2, 4i-4, 8i-8, 10i-10, 14i-14, 28i-28, 32i-32

These are all of the ns that result in primes on the axes.  When we just start looking at lots of complex integers that might yeild truly complex primes we find many more n that also appear to be unbounded.  Here are the ones I found before stopping:

i+2, 2i+1, 2i-1, i-2, -i-2, -2i-1, -2i+1, -i+2, 3i+1, 3i-1, -3i-1, -3i+1, 2i+3, 2i-3, -2i-3, -2i+3, i+4

for example, (i+2)^2 = 3+4i  yields 1+i which is prime (1^2+1^2=2) whereas (2i-1)^2=-3-4i yield -1-2i which is prime (1^2+2^2=5)

So, no pattern apparent and it looks like there are lots of answers .

  Posted by Joel on 2006-12-11 09:00:10

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