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