Earlier today, I talked about my fascination with square numbers, and their properties. Now I'm going to take it a step further and see what else we can do with square numbers.

When I was first introduced to geometry, I was immediately fascinated by the Pythagoran Theorem. This is the formula for finding the hypotenuse of a right triangle.

(Again, I'm sticking most of this in the extended section. And here's fair warning: I have never seen any of this explained anywhere else, so in many cases I've had to invent my own terms and language to describe what I'm thinking about. If anyone gets confused trying to follow the way my brain works (something that happens to me on a fairly regular basis), feel free to ask me to explain better what the hell I'm talking about. It might take me a little time to answer it, but I WILL get to you.)

For those of you unfamiliar with it, let me explain: if you have a triangle where two of the sides meet at a 90-degree angle, you can use the two shorter sides to determine the length of the third side. If A and B are the sides that meet at the right angle, then let C be the length of the third side. Simply square A and B, add them together, and take the square root of that number. Or,

A^2 + B^2 = C^2.

The simplest example is 3/4/5. Simply put, 3^2 + 4^2 = 5^2. Or, 9 + 16 = 25.

I was obsessed with this, and I started looking for more sets where all the numbers were whole numbers, not fractions or decimals. And they started popping up with almost frightening regularity. 5/12/13. 8/15/17. 7/24/25. 20/21/29. 12/35/37. 9/40/41. And so on. And so on.

But amidst the seeming randomness, I started to discern patterns.

The first one is that there is a simple way to determine whether a number can be used for C. All it has to do is fit two very simple criteria:

1) It has to be a prime number

2) When divided by 4, it has to have a remainder of .25

If it fits both rules, then it qualifies, and determining the values of A and B is simple process of elimination.

(This only works for "pure" triples, when the numbers have been simplified. I don't count 6/8/10, because it's just the 3/4/5 triple multiplied by 2. I have limited myself to prime numbers for C.)

Then I started getting "cute." What happens when you start multiplying two C's together? what sort of odd things might happen then?

Strange things. You start getting multiple solutions -- usually 4 different sets of A and B.

For example, let's start with a simple one. 5 and 13 are the first two Cs, so let's multiply them together and see what sort of As and Bs we get from 65.

The first two are obvious: 39/52/65 and 35/60/65 are simply 3/4/5 and 5/12/13, multiplied by 13 and 5 respectively. But there are two more solutions that work: 16/63/65, and 33/56/65. (I won't go into the details of how I found them, as that gets VERY strange, but feel free to test them yourself and see I'm not cheating.)

I tried it with a few more simple ones, and I started to notice a pattern: the "new" solutions always involved a multiple of the number 7. In the above example, 56 and 63 are both divisible by 7. What the hell is going on here? Where did those 7s come from?

I thought I had uncovered something very strange, almost mystical here. But I soon found that the rule was not an absolute. When I tried 5*29, or 145, the "new" solutions (24/143/145 and 17/144/145) didn't yield any new multiples of 7.

But the "old" solutions did: 20/21/29, when multiplied by 5, yields 100/105/145, and 105 is simply 7*15.

So what was the rule at play here? It seemed simple: If both or neither C had a multiple of 7 as one of its As or Bs, then the "new" solutions would have factors of 7. But if only one of them did, neither of the new solutions would. I tried it out, empirically, and it seemed to hold true.

The final step I've taken is in determining whether a C will have any 7's in its triples. There had to be some sort of universal rule for determining it, some simple test to devise that would allow me to know whether or not it would, without having to calculate the solution. And through a lot of trial and error, I found the answer:

If a prime number can be expressed as a multiple of 7 PLUS a factor of 2, then it will have a 7 in its solution. If it can only be expressed as a multiple of 7 MINUS a factor of 2, then it will not.

I tried it out on a bunch of numbers, and it seemed to hold up. 29 is 21+8, so it's good. 37 is 21+16, so it works, too. 41, though, is 42 - 1, so it doesn't. And so on. And so on.

What is the next step? I have no earthly idea. I don't know where I'll go from here, if anywhere. And I don't have enough formal math education to know if I'm on to something wholly new, something discovered centuries past, or am simply making stuff up. Part of the reason I'm tossing this out is to see if it seems familiar to anyone, who might point me towards actual scholarly work in this area. Another reason is I'm wondering if anyone else has been playing in this area, and shares my obsession.

And part of the reason is to show folks that no matter how strange you think my political opinions are, they're nothing compared to what my mathematical ideas can be.

## Comments (8)

C being 25 as in 7/24/25 is... (Below threshold)1. Posted by La Mano | January 1, 2007 12:25 PM | Score:0(0 votes cast)C being 25 as in 7/24/25 is not a prime number.

1. Posted by La Mano | January 1, 2007 12:25 PM | Score:0(0 votes cast)Posted on January 1, 2007 12:25

2. Posted by jd watson | January 1, 2007 1:00 PM | Score:0(0 votes cast)The general field of mathematics concerned with integer solutions to equations is Diophantine analysis, and this particular topic is termed Pythagorean triples.

2. Posted by jd watson | January 1, 2007 1:00 PM | Score:0(0 votes cast)Posted on January 1, 2007 13:00

3. Posted by La Mano | January 1, 2007 1:17 PM | Score:0(0 votes cast)Pythagoras Triplets are what you are fooling with.

Brahmagupta had a solution from the 7th century:

The number of triplets depends on the factors of the square of the number. A prime number will yield only one triplet, so do even numbers not evenly divisible by 4.

A^2 + B^2 = C^2

a.)

Find values of X that are factors of A^2 and A-X are a positive even number.

If A is even, the quotient A^2/X must also be even to have integral answers.

If A is odd, A^2/X must be odd. All factors of A^2 are odd so the quotient is always odd.

If A=15, possible values of X are: 9 (225/9=25), 5 (225/5=45), 3 (225/3=75, and 1 (225/1=225).

b.)

For each value of X, calculate the value of B=(A^2/X-X)/2

Thus B=(225/9-9)/2=8, or (225/5-5)/2=20, or (225/3-3)/2=36, or (225/1-1)/2=112

c.)

Also, C=B+X, or

C=8+9=17, or 20+5=25, or 36+3=39, or 112+1=113

ALL THAT YOU WANT TO KNOW IS HERE.

3. Posted by La Mano | January 1, 2007 1:17 PM | Score:0(0 votes cast)Posted on January 1, 2007 13:17

4. Posted by Jeff Michael | January 1, 2007 2:35 PM | Score:0(0 votes cast)I feel that this is kind of similar to what I've done, I dont know if anyone has found what you have done though.

One day while I was bored in math class I worked out the area of a regular polygon:

ns^2

--------

4tan(180/n)

with n being the number of sides and s being the length of a side. I proceeded to check wikipedia only to see that it was already there.

4. Posted by Jeff Michael | January 1, 2007 2:35 PM | Score:0(0 votes cast)Posted on January 1, 2007 14:35

5. Posted by Dennis Moran | January 1, 2007 3:15 PM | Score:0(0 votes cast)Did you mean to say "either rule" instead of "both rules" ?

5. Posted by Dennis Moran | January 1, 2007 3:15 PM | Score:0(0 votes cast)Posted on January 1, 2007 15:15

6. Posted by Bo | January 2, 2007 10:11 AM | Score:0(0 votes cast)Jay, your posting on mathematical matters rekindled my passion for the subject.

Somewhat relevant to this post, I hated trig until a substitute teacher revealed the unit circle basis of the functions. Don't have any idea why my regular math teacher had never introduced that, but in helping other folks with trigonometry in the years following, I get blank stares when I mention the unit circle, so I guess her omission wasn't unique.

One small question, though (and I realize you may be to busy to answer personally, so I'll pose it to any of the mathematically-inclined readers here as well): I've always felt like things like this are "hard math" while statistics and probability are "soft math," attempting to find, and in some cases creating, mathematical significance in areas that aren't directly quantifiable. In other words, am I alone in feeling that statistics is somewhat of a bastard child of mathematics?

6. Posted by Bo | January 2, 2007 10:11 AM | Score:0(0 votes cast)Posted on January 2, 2007 10:11

7. Posted by Tom Johnson | January 2, 2007 1:10 PM | Score:0(0 votes cast)YOU need to go buy a just-published book.

The book is God Created the Integers.

The editor/commentor is Stephen Hawking.

Diophantus got into square numbers big time, as did Fermat after him.

Get the book, and enjoy for the next ten years or so as you go through it.

The books is 31 mathematical discoveries, in their own words, by 17 different authors. All the usual suspects are presented: Euclid, Newton, Descartes, Gauss. In addition are some surprises: Turing, Weierstrass, Cauchy, Fourier, Diophantus, Lebesgue, Aristotle.

7. Posted by Tom Johnson | January 2, 2007 1:10 PM | Score:0(0 votes cast)Posted on January 2, 2007 13:10

8. Posted by Melinda | January 3, 2007 8:21 PM | Score:0(0 votes cast)I'm currently reading God Created the Integers. It's a tough read, but it's a lot of fun! I've seen reviews of the book which claim there are many errors in the math though, so if something doesn't make sense you may have found one of them.

I offer a correction on your rule about factors of 7 plus/minus a factor of 2. The number 41 can also be written as 35 + 6, which makes it an exception to the proposed rule.

8. Posted by Melinda | January 3, 2007 8:21 PM | Score:0(0 votes cast)Posted on January 3, 2007 20:21