
Fermat numbers and N^2+1 primes 8 messages
Hi all It now looks like that I have proven that there are only 5 fermat primes as it is conjectured to be. Furthermore I believe to have proven that there are infinetely many primes of the form N^2+1
Hi all It now looks like that I have proven that there are only 5 fermat primes as it is conjectured to be. Furthermore I believe to have proven that there are infinetely many primes of the form N^2+1

By kimowww
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4D Euler Brick does not exist
Hello Unsolved Problem Solvers, Further to my previous proof there was one counter example which has now been resolved. If the fourth parameter is even that leaves one cuboid with 3 even sides which c
Hello Unsolved Problem Solvers, Further to my previous proof there was one counter example which has now been resolved. If the fourth parameter is even that leaves one cuboid with 3 even sides which c

By Conor Williams
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Triangels with all sides prime 34 messages
Hi All I Have an idea for a geometric picture, which may turn out to look beautiful or reveal something interesting. The idea is to make a picture of triangels which have all sides prime. Make triange
Hi All I Have an idea for a geometric picture, which may turn out to look beautiful or reveal something interesting. The idea is to make a picture of triangels which have all sides prime. Make triange

By kimowww
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N^2+1 Inequality 2 messages
Hi all I have found that if the attached inequality is true, for sufficiently high values then there are infinetely many primes of the form N^2+1. If it is impossible to determine which one the sides
Hi all I have found that if the attached inequality is true, for sufficiently high values then there are infinetely many primes of the form N^2+1. If it is impossible to determine which one the sides

By kimowww
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No 3x3 Magic squares can exist (with power > 1) 2 messages
hello U.P. solvers... Further to my proof that _no_ 3x3 Magic Square of Squares can exist it follows that no other 3x3 Magic Square of _any_ Power can exist apart from power = 1. The reason being is t
hello U.P. solvers... Further to my proof that _no_ 3x3 Magic Square of Squares can exist it follows that no other 3x3 Magic Square of _any_ Power can exist apart from power = 1. The reason being is t

By Conor Williams
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Fermat the last proof with Pitagora and Binomial develop just 2 messages
Fermat the last proof with Pitagora and Binomial develop just I hope the .pdf in attachment will arrive still if heavy. I've asked to Tim for the pubblication on the wesite. Any comments is welcome. H
Fermat the last proof with Pitagora and Binomial develop just I hope the .pdf in attachment will arrive still if heavy. I've asked to Tim for the pubblication on the wesite. Any comments is welcome. H

By Stefano Maruelli
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journals
Hi all, I've recently been asked again about peerreviewed journals. A comprehensive list can be found at JOURNALS SPECIALIZING IN NUMBER THEORY. All are highly reputable. The format and submission re
Hi all, I've recently been asked again about peerreviewed journals. A comprehensive list can be found at JOURNALS SPECIALIZING IN NUMBER THEORY. All are highly reputable. The format and submission re

By Tim Roberts
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new contribution 3 messages
Hi all, There is a new contribution re the Collatz Conjecture on the UP site at http://unsolvedproblems.org/index_files/Solutions.htm It is very short and very simple, and the author would welcome fee
Hi all, There is a new contribution re the Collatz Conjecture on the UP site at http://unsolvedproblems.org/index_files/Solutions.htm It is very short and very simple, and the author would welcome fee

By Tim Roberts
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Goldbach conjecture proof ? 12 messages
Hi All It seems to me that I have found an inductive proof of Goldbach conjecture. It is pretty simple, so it should not take too long to figure out whether it is valid or not. Kim
Hi All It seems to me that I have found an inductive proof of Goldbach conjecture. It is pretty simple, so it should not take too long to figure out whether it is valid or not. Kim

By kimowww
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Perfect Number (Wave Divisor Function) 3 messages
Hello, This morning I came up with an expression for perfect numbers with help of WDF "Wave Divisor Function". See picture below: if the expression is true one has found a perfect number x. Hope I did
Hello, This morning I came up with an expression for perfect numbers with help of WDF "Wave Divisor Function". See picture below: if the expression is true one has found a perfect number x. Hope I did

By vpreemen@...
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Möbius geometry in two dimensions 5 messages
Yes. While I am not familiar with Möbius geometry in two dimensions, I am familiar with Euler's formula e^(ix) = cos(x) + i sin(x) which implies that multiplication by the complex number z = r (cos(x)
Yes. While I am not familiar with Möbius geometry in two dimensions, I am familiar with Euler's formula e^(ix) = cos(x) + i sin(x) which implies that multiplication by the complex number z = r (cos(x)

By Kermit Rose
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Goldbach conjecture 6 messages
Hi All It seems to me that I have proven that if Goldbach conjecture is false then it must have infinitely many counter examples. I wonder whether it is known if Goldbach conjecture can have only fini
Hi All It seems to me that I have proven that if Goldbach conjecture is false then it must have infinitely many counter examples. I wonder whether it is known if Goldbach conjecture can have only fini

By kimowww
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More on Division by Zero 6 messages
I have found that the solutions to 0/0=y are y=complex numbers. It is just like x°2=4, so x=2, 2. For example, 0/0=0 because 0x0=0 and 0/0=2.718 because 2.718x0=0. Danny Karl Fleming
I have found that the solutions to 0/0=y are y=complex numbers. It is just like x°2=4, so x=2, 2. For example, 0/0=0 because 0x0=0 and 0/0=2.718 because 2.718x0=0. Danny Karl Fleming

By Danny
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I'm derailed ? 2 messages
I'm derailed ? Or is it the (right) most elegant way to close the FLT proof ? It is really possible to arrive at such conclusion without performing the whole calculation (for each n) ? ... because I n
I'm derailed ? Or is it the (right) most elegant way to close the FLT proof ? It is really possible to arrive at such conclusion without performing the whole calculation (for each n) ? ... because I n

By Stefano Maruelli
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http://unsolvedproblems.org/S140.pdf 6 messages
This is in response to Enrique Santos L's email to me, who asked me to comment on his contribution here, titled "Every perfect number has only one odd prime factor". In the section titled "A simpler
This is in response to Enrique Santos L's email to me, who asked me to comment on his contribution here, titled "Every perfect number has only one odd prime factor". In the section titled "A simpler

By Jose Arnaldo BebitaDris
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new contributions
Hi all, Several new contributions recently. http://unsolvedproblems.org/index_files/Solutions.htm Tim
Hi all, Several new contributions recently. http://unsolvedproblems.org/index_files/Solutions.htm Tim

By Tim Roberts
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Factoring algorithms
I have played with my factoring program today. My goal is to see how large a number I can factor nearly instantly. This is my test run for today. Python 3.7.1 (v3.7.1:260ec2c36a, Oct 20 2018, 14:57:15
I have played with my factoring program today. My goal is to see how large a number I can factor nearly instantly. This is my test run for today. Python 3.7.1 (v3.7.1:260ec2c36a, Oct 20 2018, 14:57:15

By Kermit Rose
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Collatz Conjecture
Re:. . .The reason you switch to binary is because you are dividing by two. The holy grail is to get one bit turned on and all the rest off. That way when you divide by 2 you shift the bit to the righ
Re:. . .The reason you switch to binary is because you are dividing by two. The holy grail is to get one bit turned on and all the rest off. That way when you divide by 2 you shift the bit to the righ

By Conor Williams
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A few algebraic identities that might be useful for solving Euler box, or Perfect Cuboid, or other sum of squares puzzles.
[1] (x y z + 1)^2 – (x y z 1)^2 = (y z + x)^2 – (y z – x)^2 = (x z + y)^2 – (x z – y)^2 = (x y + z)^2 – (x y – y)^2 = 4 x y z [2] x = q1^2 p2 p3; y = p1 q2^2 p3; z = p1 p2 q3^2 [3] (p1^2 p2^2 p3^2
[1] (x y z + 1)^2 – (x y z 1)^2 = (y z + x)^2 – (y z – x)^2 = (x z + y)^2 – (x z – y)^2 = (x y + z)^2 – (x y – y)^2 = 4 x y z [2] x = q1^2 p2 p3; y = p1 q2^2 p3; z = p1 p2 q3^2 [3] (p1^2 p2^2 p3^2

By Kermit Rose
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Lonely Runner html automatic shape builder
http://shorturl.at/ltwzJ
By Conor Williams
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