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.
July 26, 2008
June 25, 2008
Confusing Myself!
I thought up a problem that I can't solve. Here it is:
Find solutions in integers n and x to the equation n^(x+2) + (n+1)^(x+1) = (n+2)^x
I found one solution, namely (n, x) = (1, 2).
However, for greater n's, the problem becomes nearly impossible to look at with number sense alone. (At any level, it can't be solved algebraically.) I've used Mathematica to determine that there aren't solutions up through n=19, and beyond there x has to be huge. Using the little number theory I know, I determined that all the solutions have to be for odd n. Beyond that, for a specific n, it is possible to find what form x has to take. (For example, when n is 9, x has to be even. This can be determined by looking at the digits value of the powers of 9 and 10, and seeing whether they sum to 1, the digits value of all powers of 11. You end up seeing that the powers of 9 are equal to 1 when x is even, and since the digits value of the powers of 10 is always 0 the sum of the LHS has the same digits value as the RHS, making solutions possible only for those values of x.)
I've also narrowed the solutions down further, and at this point I'm just trying to determine criteria that solutions will have to meet so they can be more easily determined, if they exist at all. (At this point, the outlook on finding more solutions is pretty bleak- I'm feeling pretty confident that there aren't any others- however, I've got to prove it and at the moment that looks to be a pretty difficult task.)
I thought up a problem that I can't solve. Here it is:
Find solutions in integers n and x to the equation n^(x+2) + (n+1)^(x+1) = (n+2)^x
I found one solution, namely (n, x) = (1, 2).
However, for greater n's, the problem becomes nearly impossible to look at with number sense alone. (At any level, it can't be solved algebraically.) I've used Mathematica to determine that there aren't solutions up through n=19, and beyond there x has to be huge. Using the little number theory I know, I determined that all the solutions have to be for odd n. Beyond that, for a specific n, it is possible to find what form x has to take. (For example, when n is 9, x has to be even. This can be determined by looking at the digits value of the powers of 9 and 10, and seeing whether they sum to 1, the digits value of all powers of 11. You end up seeing that the powers of 9 are equal to 1 when x is even, and since the digits value of the powers of 10 is always 0 the sum of the LHS has the same digits value as the RHS, making solutions possible only for those values of x.)
I've also narrowed the solutions down further, and at this point I'm just trying to determine criteria that solutions will have to meet so they can be more easily determined, if they exist at all. (At this point, the outlook on finding more solutions is pretty bleak- I'm feeling pretty confident that there aren't any others- however, I've got to prove it and at the moment that looks to be a pretty difficult task.)
June 1, 2008
Here's a quick and simple one.
The arithmetic average of 2kn^2 and 2k(n+1)^2 is always k greater than their geometric average for positive integers k and n.
There wasn't a uniform way to find those 2 numbers- I had to do it by finding a bunch that worked and looking for a pattern.
Note: The arithmetic average of 2 numbers a and b is (a+b)/2, and their geometric average is the Sqrt(a*b).
The arithmetic average of 2kn^2 and 2k(n+1)^2 is always k greater than their geometric average for positive integers k and n.
There wasn't a uniform way to find those 2 numbers- I had to do it by finding a bunch that worked and looking for a pattern.
Note: The arithmetic average of 2 numbers a and b is (a+b)/2, and their geometric average is the Sqrt(a*b).
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