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Old 2017-02-05, 02:45   #1
a1call
 
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Default LL Type Test for the Form 2n-3

Hi,
Is there a LL type test for numbers of the form n^2-3?
If so is it as efficient?
Thanks in advance.
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Old 2017-02-05, 02:59   #2
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Quote:
Originally Posted by a1call View Post
Hi,
Is there a LL type test for numbers of the form n^2-3?
If so is it as efficient?
Thanks in advance.
2n-3 and n^2-3 aren't the same could you clarify? We can use what we know about primes to get rid of n not needing testing for each. But I don't know if there's a LL like test for either form ( or their iterated versions). if we set 2n-3 = 6y+1 we get that 2n=6y+4 and dividing by 2 gets us n=2 mod 3. if we set 2n-3=6y+5 we get 2n=6(y+1)+2 and dividing by 2 we get n=1 mod 3 the only one of another form is n=3 leading to the prime 3. also 2n-3 is an arithmetic progression so we know how to find if any primes divide them quite quickly in one sense.

n^2-3 = 6y+1 leads to n^2 = 6y+4 which if you test squares will get you n=2,4 mod 6
n^2-3 = 6y+5 leads to n^2=6(y+1)+2 which leads to no possible result as n^2 can only be 0,1,4, or 3 mod 6
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Old 2017-02-05, 03:08   #3
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Quote:
Originally Posted by a1call View Post
Hi,
Is there a LL type test for numbers of the form n^2-3?
If so is it as efficient?
Thanks in advance.
I am sorry, I don't know what happened there.
I meant the form 2^n-3.
I thought that's what I typed in the title as well as the OP.
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Old 2017-02-05, 03:32   #4
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Quote:
Originally Posted by a1call View Post
I am sorry, I don't know what happened there.
I meant the form 2^n-3.
I thought that's what I typed in the title as well as the OP.
I think we can put it in the correct form to do one of the tests listed here ( even if it might take Proth's theorem). but I don't know if it would be efficient.
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Old 2017-02-05, 18:40   #5
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Quote:
Originally Posted by a1call View Post
Hi,
Is there a LL type test for numbers of the form n^2-3 2^n-3?
No.
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Old 2017-02-05, 19:05   #6
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Thank you.
Just the sort of answer I appreciate for Yes or No questions.

Last fiddled with by a1call on 2017-02-05 at 19:06
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Old 2017-02-05, 20:43   #7
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Originally Posted by Batalov View Post
No.
really not even LLR ? I would get something like (2^n-1)*2^1+1 for example though I guess that's too simple an attempt at an actual answer. edit: doh idiocy alert (2^n-1)*(2^1)-1 which makes it a proth theorem prime according the LLR wikipedia.

Last fiddled with by science_man_88 on 2017-02-05 at 21:18
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Old 2017-02-05, 21:36   #8
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Quote:
Originally Posted by science_man_88 View Post
really not even LLR ? I would get something like (2^n-1)*2^1+1 for example though I guess that's too simple an attempt at an actual answer. edit: doh idiocy alert (2^n-1)*(2^1)-1 which makes it a proth theorem prime according the LLR wikipedia.
Would you please care to read the Proth theorem's first flipping line?
What is the relationship between k and 2n for the Proth theorem to be applicable?
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Old 2017-02-05, 21:40   #9
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Quote:
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Would you please care to read the Proth theorem's first flipping line?
What is the relationship between k and 2n for the Proth theorem to be applicable?
okay I'll admit I didn't read https://en.wikipedia.org/wiki/Lucas%...%93Riesel_test closely enough while trying to figure out stuff for myself amybe this si why I don't google things before posting ( like RDS would suggest). if my math knowledge isn't high enough to begin with then there's clearly no google search I can do that would help.
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Old 2017-02-05, 22:14   #10
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okay I'll admit I didn't read https://en.wikipedia.org/wiki/Lucas%...%93Riesel_test closely enough while trying to figure out...
Did you read the first line of Riesel's Thm, then? what is the relationship of k and 2n?
I am not asking you read "closely enough". I am asking to read the first line.
Hint: the first line of both theorems clearly says k < 2^n, doesn't it?
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Old 2017-02-05, 22:18   #11
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Quote:
Originally Posted by Batalov View Post
Did you read the first line of Riesel's Thm, then? what is the relationship of k and 2n?
I am not asking you read "closely enough". I am asking to read the first line.
Hint: the first line of both theorems clearly says k < 2^n, doesn't it?
yes I guess they do.
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