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## Re: [PrimeNumbers] Another form of probabilistic test for primeness

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• ... Four shalt be the number thou shalt count, and the number of the counting shalt be four. For any odd prime x, we have 2^(4x+1) = 2^(5 + 4(x-1)), so
Message 1 of 3 , Jun 23, 2012
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> It will be a matter of empirical research to determine lower bounds
> on composite y such that
>
> 2^(3y+1) = 2^(y+3) mod (3y)
> 2^(5y+1) = 2^(y+5) mod (5y)
> 2^(7y+1) = 2^(y+7) mod (7y)

Four shalt be the number thou shalt count, and the number of the counting
shalt be four.

For any odd prime x, we have 2^(4x+1) = 2^(5 + 4(x-1)), so
2^(4x+1) = 0 (mod 4) and 2^(4x+1) = 2^5 (mod x), thus by Chinese Remainder
Theorem, we get 2^(4x+1) = 2^5 (mod 4x)

We also have 2^(x+4) = 2^(5 + (x-1)), which gets us
2^(x+4) = 0 (mod 4) and 2^(x+4) = 2^5 (mod x). Combining them by CNT
gives 2^(x+4) = 2^5 (mod 4x)

Ergo, 2^(4x+1) = 2^(4+x) mod (4x), and 4 is the smallest counterexample
for any odd prime x.

Peter
• ... counting shalt be four. ... 0 (mod 4) and 2^(4x+1) = 2^5 (mod x), thus by Chinese Remainder Theorem, we get 2^(4x+1) = 2^5 (mod 4x) ... counterexample for
Message 2 of 3 , Jun 24, 2012
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On 6/23/2012 5:44 PM, Peter Kosinar wrote:
>> It will be a matter of empirical research to determine lower bounds
>> on composite y such that
>>
>> 2^(3y+1) = 2^(y+3) mod (3y)
>> 2^(5y+1) = 2^(y+5) mod (5y)
>> 2^(7y+1) = 2^(y+7) mod (7y)
>
> Four shalt be the number thou shalt count, and the number of the
counting shalt be four.
>
> For any odd prime x, we have 2^(4x+1) = 2^(5 + 4(x-1)), so 2^(4x+1) =
0 (mod 4) and 2^(4x+1) = 2^5 (mod x), thus by Chinese Remainder Theorem,
we get 2^(4x+1) = 2^5 (mod 4x)
>
> We also have 2^(x+4) = 2^(5 + (x-1)), which gets us
> 2^(x+4) = 0 (mod 4) and 2^(x+4) = 2^5 (mod x). Combining them by CNT
> gives 2^(x+4) = 2^5 (mod 4x)
>
> Ergo, 2^(4x+1) = 2^(4+x) mod (4x), and 4 is the smallest
counterexample for any odd prime x.
>
> Peter

Hello Peter.

Thanks.

My test is even worse than you have discovered.

For any composite y such that 2^y = 2 mod y, (Original Fermat test
which fails too many times)
and for any odd prime x,

Calculate 2^(xy-x-y+1) mod x and 2^(xy-x-y+1) mod y

2^(xy-x-y+1) = (2^x)^y * 2^(-x-y+1) = 2^y * 2^(-x-y+1) = 2^(y-x-y+1) =
2^(-x+1) = 2^(1-1) mod x

2^(xy-x-y+1) = (2^x)^y * 2^(-x-y+1) = 2^x * 2^(-x-y+1) = 2^(x-x-y+1) =
2^(-y+1) = 2^(1-1) mod y

Thus by chinese remainder theorem, 2(xy-x-y+1) = 1 mod (xy).

This proves that my proposed "new" test is only a disguised Fermat test.

Kermit Rose
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