larryfreeman
May8-11, 04:33 AM
I recently came across an argument that seems to prove Legendre's Conjecture. Since I am a mathematical amateur, I am sending my posting to this forum in hopes that someone here will help me to quickly find my flaw.
The argument is very simple which indicates that it has a flaw. Here's the link for the full proof:
http://hubpages.com/hub/A-Proposed-Proof-for-Legendres-Conjecture
Below I try to present a summary of my basic argument.
The argument is based on the observation that if a sequence of consecutive integers is of length n and n+1 is a prime, then at least one of the integers in the sequence is not divisible by any prime less than n.
If x is any integer, by sequence of consecutive integers, I mean x+1, x+2, x+3 ..., x+n.
Unfortunately, this argument hinges on three concepts which are not standard mathematical ideas. I will be very glad to remove these definitions if someone can help me formulate the ideas in standard mathematical terms.
The ideas are what I am calling:
(1) Sequence Divisibility Problem: Are all the integers in a sequence of n consecutive integers divisible by a prime less than or equal to n (for example, it is solvable for n=3 but not solvable for n=4)
In other words, does an integer x exist such that x+1, x+2, ..., x+n has the property that all integers in this sequence are divisible by a prime less than or equal to n.
(2) n-Slot Form: A list of primes where slot 1 corresponds to the first integer in a sequence, slot 2 corresponds to the second integer in the sequence etc up to slot n.
If a sequence divisibility problem has a solution then it is possible to fill each slot in an n-Slot Form with a prime less than or equal to n.
(3) Ascending Prime Order Form: This is an n-Slot Form where each prime p is first placed in the p-1 slot and as you can tell from this assumption, the first slot that a prime appears is in ascending prime order.
I apologize for these three ideas. I know I hate it when an amateur presents all these strange terms and then proceeds to make giant leaps to get to the conclusion of a proof.
The basic argument in my proof is the following:
(1) If a solution to Sequence Divisibility Problem exists, then the solution is expressible in Ascending Prime Order Form.
(2) If a sequence of consecutive integers is of length n and n+1 is a prime, then it is not possible to express a solution in Ascending Prime Order Form.
(3) Therefore, it follows that no solution exists and there must be at least one integer in the sequence that is not divisible by a prime less than or equal to n.
(4) From this conclusion and Bertrand's Postulate, I attempt to show by proof-by-contradiction that Legendre's Conjecture must be true.
I hope this summary helps clarify my approach. I hope that it sounds reasonable enough that you can help me find the flaw or have suggestions for refinement so that the proof is easier to follow without exerting too much effort.
Thanks very much for your time if you have read this far! :-)
-Larry
Update: The flaw in my reasoning is Lemma 3. Lemma 3 is wrong. A sequence divisibility problem can be solvable without having a solution in ascending prime order form.
This is shown by n=22. In this case, the
Sequence Divisibility Problem has a solution (1->2, 2->5, 4->3, 6->7,
8->11, 14->13, 18->17) but there is no solution in Ascending Prime
Form.
We can use the Chinese Remainder Theorem to solve for:
x+1 = 0 (mod 2)
x+1 = -1 (mod 5)
x+1 = 0 (mod 3)
x+1 = -5 (mod 7)
x+1 = -7 (mod 11)
x+1 = 0 (mod 13)
x+1 = 0 (mod 17)
But there is no sequence in ascending prime form that has a solution.
I guess that I missed this because I would have expected the flaw to
appear elsewhere. My arguments seems to be valid for assuming that 2
is in the first slot but it is invalid when trying to generalize this
log to 3 or for that matter to any prime p.
The argument is very simple which indicates that it has a flaw. Here's the link for the full proof:
http://hubpages.com/hub/A-Proposed-Proof-for-Legendres-Conjecture
Below I try to present a summary of my basic argument.
The argument is based on the observation that if a sequence of consecutive integers is of length n and n+1 is a prime, then at least one of the integers in the sequence is not divisible by any prime less than n.
If x is any integer, by sequence of consecutive integers, I mean x+1, x+2, x+3 ..., x+n.
Unfortunately, this argument hinges on three concepts which are not standard mathematical ideas. I will be very glad to remove these definitions if someone can help me formulate the ideas in standard mathematical terms.
The ideas are what I am calling:
(1) Sequence Divisibility Problem: Are all the integers in a sequence of n consecutive integers divisible by a prime less than or equal to n (for example, it is solvable for n=3 but not solvable for n=4)
In other words, does an integer x exist such that x+1, x+2, ..., x+n has the property that all integers in this sequence are divisible by a prime less than or equal to n.
(2) n-Slot Form: A list of primes where slot 1 corresponds to the first integer in a sequence, slot 2 corresponds to the second integer in the sequence etc up to slot n.
If a sequence divisibility problem has a solution then it is possible to fill each slot in an n-Slot Form with a prime less than or equal to n.
(3) Ascending Prime Order Form: This is an n-Slot Form where each prime p is first placed in the p-1 slot and as you can tell from this assumption, the first slot that a prime appears is in ascending prime order.
I apologize for these three ideas. I know I hate it when an amateur presents all these strange terms and then proceeds to make giant leaps to get to the conclusion of a proof.
The basic argument in my proof is the following:
(1) If a solution to Sequence Divisibility Problem exists, then the solution is expressible in Ascending Prime Order Form.
(2) If a sequence of consecutive integers is of length n and n+1 is a prime, then it is not possible to express a solution in Ascending Prime Order Form.
(3) Therefore, it follows that no solution exists and there must be at least one integer in the sequence that is not divisible by a prime less than or equal to n.
(4) From this conclusion and Bertrand's Postulate, I attempt to show by proof-by-contradiction that Legendre's Conjecture must be true.
I hope this summary helps clarify my approach. I hope that it sounds reasonable enough that you can help me find the flaw or have suggestions for refinement so that the proof is easier to follow without exerting too much effort.
Thanks very much for your time if you have read this far! :-)
-Larry
Update: The flaw in my reasoning is Lemma 3. Lemma 3 is wrong. A sequence divisibility problem can be solvable without having a solution in ascending prime order form.
This is shown by n=22. In this case, the
Sequence Divisibility Problem has a solution (1->2, 2->5, 4->3, 6->7,
8->11, 14->13, 18->17) but there is no solution in Ascending Prime
Form.
We can use the Chinese Remainder Theorem to solve for:
x+1 = 0 (mod 2)
x+1 = -1 (mod 5)
x+1 = 0 (mod 3)
x+1 = -5 (mod 7)
x+1 = -7 (mod 11)
x+1 = 0 (mod 13)
x+1 = 0 (mod 17)
But there is no sequence in ascending prime form that has a solution.
I guess that I missed this because I would have expected the flaw to
appear elsewhere. My arguments seems to be valid for assuming that 2
is in the first slot but it is invalid when trying to generalize this
log to 3 or for that matter to any prime p.