Is it possible to find a proper divisor of a composite Fermat Number without using any arithmetic operation...
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I have an absurdly simple algorithm which for N = 2^{2^{k-1}} +1 a composite Fermat Number,
computes a divisor 1
The algorithm is also a test for primality, i.e. N is prime if and only if my algorithm
has output e=N.
Features:
The algorithm is a k-step recursion. Hence it contains no steps which involve "Trial & Error".
The algorithm involves simple manipulations of bit strings. It uses non of the
arithmetic operations +,-,*,/.
My goal is to prove that no such algorithm as I have just described is possible. This would be the contradiction in a theorem I proving.
number-theory
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add a comment |
$begingroup$
I have an absurdly simple algorithm which for N = 2^{2^{k-1}} +1 a composite Fermat Number,
computes a divisor 1
The algorithm is also a test for primality, i.e. N is prime if and only if my algorithm
has output e=N.
Features:
The algorithm is a k-step recursion. Hence it contains no steps which involve "Trial & Error".
The algorithm involves simple manipulations of bit strings. It uses non of the
arithmetic operations +,-,*,/.
My goal is to prove that no such algorithm as I have just described is possible. This would be the contradiction in a theorem I proving.
number-theory
$endgroup$
2
$begingroup$
All of those arithmetic operations can be implemented as simple manipulations of bit strings.
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– user3482749
Nov 24 '18 at 17:36
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Proving in general that an algorithm doesn't exist is extremely difficult. For example, the following question is open: Is there are a linear time algorithm for multiplying two positive integers?
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– JoshuaZ
Nov 24 '18 at 17:57
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Sure, arithmetic operations can be implemented as simple manipulations on bit strings. But with my algorithm the point is that none of what I do amounts to performing any arithmetic operations.
$endgroup$
– Arthur Lubocce
Nov 26 '18 at 5:25
$begingroup$
What I am attempting to show is the opposite of finding some optimal algorithm. I seek to prove that all algorithms to find divisors of Fermat Numbers or to prove they are composite requires some minimal operations, namely arithmetic operations.
$endgroup$
– Arthur Lubocce
Nov 26 '18 at 5:28
add a comment |
$begingroup$
I have an absurdly simple algorithm which for N = 2^{2^{k-1}} +1 a composite Fermat Number,
computes a divisor 1
The algorithm is also a test for primality, i.e. N is prime if and only if my algorithm
has output e=N.
Features:
The algorithm is a k-step recursion. Hence it contains no steps which involve "Trial & Error".
The algorithm involves simple manipulations of bit strings. It uses non of the
arithmetic operations +,-,*,/.
My goal is to prove that no such algorithm as I have just described is possible. This would be the contradiction in a theorem I proving.
number-theory
$endgroup$
I have an absurdly simple algorithm which for N = 2^{2^{k-1}} +1 a composite Fermat Number,
computes a divisor 1
The algorithm is also a test for primality, i.e. N is prime if and only if my algorithm
has output e=N.
Features:
The algorithm is a k-step recursion. Hence it contains no steps which involve "Trial & Error".
The algorithm involves simple manipulations of bit strings. It uses non of the
arithmetic operations +,-,*,/.
My goal is to prove that no such algorithm as I have just described is possible. This would be the contradiction in a theorem I proving.
number-theory
number-theory
asked Nov 24 '18 at 16:58
Arthur LubocceArthur Lubocce
1
1
2
$begingroup$
All of those arithmetic operations can be implemented as simple manipulations of bit strings.
$endgroup$
– user3482749
Nov 24 '18 at 17:36
$begingroup$
Proving in general that an algorithm doesn't exist is extremely difficult. For example, the following question is open: Is there are a linear time algorithm for multiplying two positive integers?
$endgroup$
– JoshuaZ
Nov 24 '18 at 17:57
$begingroup$
Sure, arithmetic operations can be implemented as simple manipulations on bit strings. But with my algorithm the point is that none of what I do amounts to performing any arithmetic operations.
$endgroup$
– Arthur Lubocce
Nov 26 '18 at 5:25
$begingroup$
What I am attempting to show is the opposite of finding some optimal algorithm. I seek to prove that all algorithms to find divisors of Fermat Numbers or to prove they are composite requires some minimal operations, namely arithmetic operations.
$endgroup$
– Arthur Lubocce
Nov 26 '18 at 5:28
add a comment |
2
$begingroup$
All of those arithmetic operations can be implemented as simple manipulations of bit strings.
$endgroup$
– user3482749
Nov 24 '18 at 17:36
$begingroup$
Proving in general that an algorithm doesn't exist is extremely difficult. For example, the following question is open: Is there are a linear time algorithm for multiplying two positive integers?
$endgroup$
– JoshuaZ
Nov 24 '18 at 17:57
$begingroup$
Sure, arithmetic operations can be implemented as simple manipulations on bit strings. But with my algorithm the point is that none of what I do amounts to performing any arithmetic operations.
$endgroup$
– Arthur Lubocce
Nov 26 '18 at 5:25
$begingroup$
What I am attempting to show is the opposite of finding some optimal algorithm. I seek to prove that all algorithms to find divisors of Fermat Numbers or to prove they are composite requires some minimal operations, namely arithmetic operations.
$endgroup$
– Arthur Lubocce
Nov 26 '18 at 5:28
2
2
$begingroup$
All of those arithmetic operations can be implemented as simple manipulations of bit strings.
$endgroup$
– user3482749
Nov 24 '18 at 17:36
$begingroup$
All of those arithmetic operations can be implemented as simple manipulations of bit strings.
$endgroup$
– user3482749
Nov 24 '18 at 17:36
$begingroup$
Proving in general that an algorithm doesn't exist is extremely difficult. For example, the following question is open: Is there are a linear time algorithm for multiplying two positive integers?
$endgroup$
– JoshuaZ
Nov 24 '18 at 17:57
$begingroup$
Proving in general that an algorithm doesn't exist is extremely difficult. For example, the following question is open: Is there are a linear time algorithm for multiplying two positive integers?
$endgroup$
– JoshuaZ
Nov 24 '18 at 17:57
$begingroup$
Sure, arithmetic operations can be implemented as simple manipulations on bit strings. But with my algorithm the point is that none of what I do amounts to performing any arithmetic operations.
$endgroup$
– Arthur Lubocce
Nov 26 '18 at 5:25
$begingroup$
Sure, arithmetic operations can be implemented as simple manipulations on bit strings. But with my algorithm the point is that none of what I do amounts to performing any arithmetic operations.
$endgroup$
– Arthur Lubocce
Nov 26 '18 at 5:25
$begingroup$
What I am attempting to show is the opposite of finding some optimal algorithm. I seek to prove that all algorithms to find divisors of Fermat Numbers or to prove they are composite requires some minimal operations, namely arithmetic operations.
$endgroup$
– Arthur Lubocce
Nov 26 '18 at 5:28
$begingroup$
What I am attempting to show is the opposite of finding some optimal algorithm. I seek to prove that all algorithms to find divisors of Fermat Numbers or to prove they are composite requires some minimal operations, namely arithmetic operations.
$endgroup$
– Arthur Lubocce
Nov 26 '18 at 5:28
add a comment |
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2
$begingroup$
All of those arithmetic operations can be implemented as simple manipulations of bit strings.
$endgroup$
– user3482749
Nov 24 '18 at 17:36
$begingroup$
Proving in general that an algorithm doesn't exist is extremely difficult. For example, the following question is open: Is there are a linear time algorithm for multiplying two positive integers?
$endgroup$
– JoshuaZ
Nov 24 '18 at 17:57
$begingroup$
Sure, arithmetic operations can be implemented as simple manipulations on bit strings. But with my algorithm the point is that none of what I do amounts to performing any arithmetic operations.
$endgroup$
– Arthur Lubocce
Nov 26 '18 at 5:25
$begingroup$
What I am attempting to show is the opposite of finding some optimal algorithm. I seek to prove that all algorithms to find divisors of Fermat Numbers or to prove they are composite requires some minimal operations, namely arithmetic operations.
$endgroup$
– Arthur Lubocce
Nov 26 '18 at 5:28