Richey Numbers: An Overview
After reviewing this problem further for a while, I came to some new conclusions.
First, (I stole this one from a book by Paul J. Nahin), there is an alternate way to generate numbers of the form a^2+b^2 = c^2+d^2. His method uses the product of the sum of two squares:
(u^2+v^2)(x^2+y^2)
The theorem says that a product of this form (for integral values u, v, x and y) will always be equal to a number that can be expressed in two distinct ways as the sum of two squares. This can be verified by some algebra and the solutions can be found in terms of u, v, x and y.
This, along with a new ability I have maintained to use C++ code to find solutions to some of these number theory problems, led me to write a program that found solutions, and it works quite well. This leads me to another, quicker conclusion. The smallest Richey number that is not a multiple of 5 is 221, not 533 as I had said earlier. (221 = 5^2+14^2 = 10^2 + 11^2)
Finally, there is another element of these numbers that I have not yet studied. There appear some relationships between the numbers which compose my numbers. One example, which I noted last time, was the relationship that a+b-c-d was always an even number. There appear some more of these strange relationships, but I haven't yet determined anything with proof.
December 9, 2008
September 9, 2008
Taxicab Numbers Extended: Extended
I liked my extension of taxicab numbers so much that I've composed an addition to it.
These three formulas are ones I came up with to generate an infinite number of the numbers I talked about last time.
5(2n^2+2n+1)
5(n^2+4n+5)
13(2n^2+10n+13)
The first only works for n greater than or equal two, whereas the other two work for positive integral n.
These formulas were found by looking at the problem as a difference of squares problem, and from there as this: Finding two strings of consecutive odd integers that are equal.
The third formula also allowed me to disprove one of my earlier conjectures; that all of the numbers are multiples of 5. This formula found a whole set of them, the smallest of which is 533. (23^2 + 2^2 and 22^2 + 7^2.)
There is still more to do, since I haven't yet found a formula that will determine every single number of this type, though having an infinite number of them is, well, pretty damn cool.
I liked my extension of taxicab numbers so much that I've composed an addition to it.
These three formulas are ones I came up with to generate an infinite number of the numbers I talked about last time.
5(2n^2+2n+1)
5(n^2+4n+5)
13(2n^2+10n+13)
The first only works for n greater than or equal two, whereas the other two work for positive integral n.
These formulas were found by looking at the problem as a difference of squares problem, and from there as this: Finding two strings of consecutive odd integers that are equal.
The third formula also allowed me to disprove one of my earlier conjectures; that all of the numbers are multiples of 5. This formula found a whole set of them, the smallest of which is 533. (23^2 + 2^2 and 22^2 + 7^2.)
There is still more to do, since I haven't yet found a formula that will determine every single number of this type, though having an infinite number of them is, well, pretty damn cool.
July 26, 2008
Taxicab Numbers Extended
Taxicab numbers are numbers that can be written in two distinct ways as the sum of two cubes, the smallest of which is 1729. (1729 = 1^3 + 12^3 = 9^3 + 10^3.)
I decided it would be a good idea to try and find numbers that can be written as the sum of two squares in two distinct ways, and so far I have found 18 of them. In ascending order,
50, 65, 85, 125, 130, 145, 170, 185, 205, 250, 260, 265, 290, 305, 325, 340, 365, 370.
Unfortunately, I haven't determined a uniform way to find these numbers- so, I've just been checking all the multiples of 5. (I checked every number between 1 and 100, and noticing that numbers that satisfy k = 0 mod 5 are the only ones that work, I only checked multiples of 5 from then on.) So, there could be some missing in between. (There might even be some multiples of 5 I missed in the 300's because I kind of got lazy and started writing down the ones that worked and not checking the ones I was pretty sure didn't work.) I also have no idea why only numbers that are multiples of 5 would work, but I have a vague idea.
If we look at the digit that ends up in the digits place when you square the 10 different 1 digit numbers, we get 0, 1, 4, 9, 6, 5, 6, 9, 4, 1. There are 6 digits you can end up with, namely 0, 1, 4, 5, 6 and 9. Combining these, you can make any digit. However, most can only be made with one or two combinations of numbers- 5 and 0 can be made with 3 different combinations. (5 = 1+4 = 0+5 = 9+6, and 0 = 0+0 = 1+9 = 4+6.) This makes these two digits the most likely to be in the digits place, but it is no help in proving that they will be the only digits in the digits place of my numbers.
Another more obscure pattern that I noticed is in the numbers that you square to create my numbers. For example, 50 = 5^2 + 5^2 = 7^2 + 1^2. Notice that 5+5 = 10 is 2 greater than 7+1 = 8. In other words, for a number k that can be written as both a^2 + b^2 and c^2 + d^2 for positive integers a, b, c and d and for a+b > c+d, a+b-c-d = 2. However, this only holds for the first 5- after that, the difference jumps to either 4 or 6 or back to 2. Although less specific, another rule can form- perhaps a+b-c-d = 0 mod 2.
The numbers seem to show up almost randomly, and the problem is a relatively unfamiliar one so I'm not sure where to go from here.
Taxicab numbers are numbers that can be written in two distinct ways as the sum of two cubes, the smallest of which is 1729. (1729 = 1^3 + 12^3 = 9^3 + 10^3.)
I decided it would be a good idea to try and find numbers that can be written as the sum of two squares in two distinct ways, and so far I have found 18 of them. In ascending order,
50, 65, 85, 125, 130, 145, 170, 185, 205, 250, 260, 265, 290, 305, 325, 340, 365, 370.
Unfortunately, I haven't determined a uniform way to find these numbers- so, I've just been checking all the multiples of 5. (I checked every number between 1 and 100, and noticing that numbers that satisfy k = 0 mod 5 are the only ones that work, I only checked multiples of 5 from then on.) So, there could be some missing in between. (There might even be some multiples of 5 I missed in the 300's because I kind of got lazy and started writing down the ones that worked and not checking the ones I was pretty sure didn't work.) I also have no idea why only numbers that are multiples of 5 would work, but I have a vague idea.
If we look at the digit that ends up in the digits place when you square the 10 different 1 digit numbers, we get 0, 1, 4, 9, 6, 5, 6, 9, 4, 1. There are 6 digits you can end up with, namely 0, 1, 4, 5, 6 and 9. Combining these, you can make any digit. However, most can only be made with one or two combinations of numbers- 5 and 0 can be made with 3 different combinations. (5 = 1+4 = 0+5 = 9+6, and 0 = 0+0 = 1+9 = 4+6.) This makes these two digits the most likely to be in the digits place, but it is no help in proving that they will be the only digits in the digits place of my numbers.
Another more obscure pattern that I noticed is in the numbers that you square to create my numbers. For example, 50 = 5^2 + 5^2 = 7^2 + 1^2. Notice that 5+5 = 10 is 2 greater than 7+1 = 8. In other words, for a number k that can be written as both a^2 + b^2 and c^2 + d^2 for positive integers a, b, c and d and for a+b > c+d, a+b-c-d = 2. However, this only holds for the first 5- after that, the difference jumps to either 4 or 6 or back to 2. Although less specific, another rule can form- perhaps a+b-c-d = 0 mod 2.
The numbers seem to show up almost randomly, and the problem is a relatively unfamiliar one so I'm not sure where to go from here.
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