What's new (July 22, 2009)
Chapter 0 : Opening  Why I am interested in Number Theory.
Chapter 1 : 4/n = 1/a + 1/b + 1/c [D11]
Chapter 2 : Squares consisted of 3 different digits [F24]
Chapter 3 : n = (x + y + z)(1/x + 1/y + 1/z)
Chapter 4 : n = x^{3} + y^{3} + z^{3} [D5]
Chapter 5 : Repeating Decimals [A3]
Chapter 6 : Additive Palindromicness of Natural Numbers [F32]
Chapter 7 : Collatz's Conjecture [E16]
Chapter 8 : Continued Fraction and Pell's Equation
Chapter 9 : Amicable Numbers [B4][B5]
Chapter 10 : Congruent Numbers (Congruum) [D27]
Chapter 11 : Number Theoretic Algorithms [A3] [B45] [F10]
Chapter 12 : Integer Factorization Algorithms
Appendix 1 : WIFC (World Integer Factorization Center)
Appendix 2 : Bibliography
Appendix 3 : How to join in the Factorization Project
Appendix 4 : Benchmark of Tomabechi's "ppmpqs.exe"
Appendix 5 : (new !) Prime table (Primes near to 2^{n}, 10^{n}, factorial, primorial and compositorial)
[nn] denote the number in Richard K. Guy,
"Unsolved Problems in Number Theory"
Third Edition.
(A few contents are still written in Japanese)
Erdös and Strauss conjectured that the following Diopantine equation
4/n = 1/a + 1/b + 1/ccould be solved in positive integers for all n < 1.
I found how to construct the parameterize solution from arbitrary solutions.
If only C is divisable by P, then there is another way to construct the parameterize solution heuristically.
Theorem :
Let A, B, C in N be a solution of the Diophantine equation
m/P=1/A+1/B+1/C, B=kP(Notice : C is always divisable by P)
(A<B<C; m=4, 5, 6, 7; P=prime; k∈N)
Define a, b, c, d, e, f, c', d' as,c := B/PThen,
a := mk1
b := a(P mod a)
n := (P+b)/a
d := cnA
e := gcd(c,d)
c':= c/e
d':= d/e
f := ke/(bcad)
m/(anb)=1/e(c'nd')+1/k(anb)+1/f(anb)(c'nd')andP = anb
A = e(c'nd')
B = k(anb)
C = f(anb)(c'nd)
4/P=1/A+1/B+1/C : P=2..47, P=53..97
5/P=1/A+1/B+1/C : P=2..47, P=53..97
6/P=1/A+1/B+1/C : P=2..47, P=53..97
7/P=1/A+1/B+1/C : P=2..47, P=53..97
Japanese mathematician Sin Hitotsumatsu asked the proof or the contradiction that, apart from tribial solutions 10^{2n}, 4*10^{2n} and 9*10^{2n}, there are perfect square number consisted of only 2 different decimal digits. Known largest solution is
81619^{2} = 6661661161.
I extended the condition of this problem as (i) 3 different digits (ii) perfect nth power,
and found many remarkable parametrized patterns and sporadic solutions.
All solutions of sporadic pattern up to 10^{25} for nonzero patterns
and up to 10^{23} patterns including zero are
here.
150167406766664999985^{2}= 22550250055025025225250200022000050000225 (April 07, 2008)All solutions of infinite pattern (parameterized solutions) are here.
149067065510873088673^{2}= 22220990020022929092929022220290920900929 (April 07, 2008)
3180252254777039538502^{2}= 10114004404014444004140001011411401140404004 (April 28, 2008)
6674983479713230005962^{2}= 44555404454444540454555540045000554555545444 (June 02, 2008)
3015775265159011230138^{2}= 9094900449944904494440090444449999999499044 (June 09, 2008)
44949994999999949999995^{2}= 2020502050500020505000050500052500000500000025 (October 31, 2008)
20832739723817975138362^{2}= 434003044400343443044430000434430333044043044 (October 31, 2008)
Sometimes the first (smallest) solution becomes very large. For example,
056 : 2236081408416666^{2} = 5000060065066660656065066555556 (May 4, 1997)
079 : 8819172285373497^{2} = 77777799799099990007000790009009 (May 5, 1997)
789 : 9949370777987917^{2} = 98989978877879888789778997998889 (May 10, 1997)
019 : 43694278824566964251^{2} = 1909190001999001011109190090109911991001 (May 6, 1998)
I searched up to the following range, but couldn't find the solution of 013 and 689;
013 : 10^{24}
678 : 10^{25}
The solutions for higher powers are here.
I couldn't find the perfect 7th power consisted of 3 different digits.
Searching for integer solutions of above equation for abs(n) is less than 100 by primitive way.
Solutions 100 ≤ n ≤ 100 (by exhaustive search) are here.
Solutions with Cremona's mwrank
Solutions 500 ≤ n ≤ 1Now you can get larger solutions like
Solutions 1 ≤ n ≤ 500
n = 100 : (x,y,z) = (4450012553, 219887106322, 663397965750)
n = 94 : (x,y,z) = (571064, 1799160, 79045681)
Searching for integer solutions of the equation x/y+y/z+z/x=n
for abs(n) is less than 100 by primitive way.
Solutions 100 ≤ n ≤ 100 (by exhaustive search) are here.
Solutions with Cremona's mwrank
Solutions for 100 ≤ n ≤ 1Now you can get larger solutions like
Solutions for 1 ≤ n ≤ 100
Searching for above equation's integer solutions of n less than 10000.
Solutions are here.
Search range is now max { x , y , z } ≤ 10^{6}.
For certain numbers, search range is up to 10^{10} and beyond,
which achieved by Dan J. Bernstein and Noam D. Elkies.
And solutions of n = x^{3} + y^{3} + 2z^{3} are here.
Search range is now max { x , y , z } ≤ 10^{6}.
JeanCharles Meyrignac and Mike Oakes tried up to max { x , y , z } ≤ 4*10^{6}.
Introduction of Repeating Decimals (this is not an unsolved problem).
Also referring to factorizations of Repunits
(A3 Mersenne primes. Repunits. Fermat numbers. Primes of shape k.2^{n}+1).
(A few contents are still written in Japanese)
Almost all nutural number seems to become palindromic after several iteration of addition with reverse order of itself.
196 is not known whether it will become palindromic or not.
And there are some numbers which will become palindromic after several iteration of
addition but it takes many times. For example, 89 requires 24 times iterations.
The largest known iteration times is 10000000525586206 (232 times) found by Helmut Postl.
Other examples are listed here.
This is very famous unsolved problem.
For n∈N, define f(n) as,
3n+1 (n is odd)Then any n (seems to) converge to the sequence 4  2  1.
n/2 (n is even).
I suggest how to construct the apparantly convergent equation like 4n+1.
(4n+1  (12n+3)+1  3n+1, (3n+1) < (4n+1) )
By using this method, we can reduce the searching range.
The ratio are shown
here.
I also consider the relation between n and max(n) (largest value of f(n)).
The table is
here.
It seems to be that there exists constant s (almost equal to 2) such that
max(n) ≤ n^{2+ε}If it is proven, it would be the upper bound of f(n) for arbitrary n,
Diophantine equation x^{2}ny^{2}=±1
is called Pell equation.
Sometimes x and y becomes large for some n. For example,
227528^{2}  103 * 22419^{2} = 1
8890182^{2}  109 * 851525^{2} = 1
We can solve Pell's Equation using continued fraction expansion.
A program is here.
(this is not an unsolved problem).
Searching the following number's pairs up to 10^{10}.
Amicable numbers
Quasiamicable numbers
Augmented amicable numbers.
(A few contents are still written in Japanese)
If n∈N is the value of area of rightangled triangle and
the values of each sides of triangle are in Q, then n is called as
congruent number or congruum.
How to construct the solution of n is not known.
I showed some construction method for the solutions of congruum.
The solutions of congruum up to 1000 are
here
and locate here.
(A few contents are still written in Japanese)
Introduction of various number theoretic algorithms.
Following topics are rarely seen in the elementary course, I think.
Wieferich prime (A3 Mersenne primes. Repunits. Fermat numbers. Primes of shape k.2^{n}+1)
(A few contents are still written in Japanese)
Following algorithms are mentioned, and also shown the sample programs.
including the factors of following problem;
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