NOTE: I'm very very sorry if this has already been done before and due to my ignorance is very very boring.

Okay, let a pythagorean set be a,b,c.

There are many types of pythagorean triplets, but essentially, they all comed down to having one step between the second and third number, two steps, three steps, and so on. There are not crazy jumps in the difference between b and c, which makes my idea work.

Example: 3,4,5 is a "one-step" set because 5-4 is one. 8,15,17 is a two step set because 17-15 is 2. Simple enough.

So here's what I did:

a^2 + b^2 = c^2

c^2 - b^2 = a^2

The next step assumes that a,b,c in this case is a "one-step" set.

c^2 - (c-1)^2 = a^2

c^2 -c^2 + 2c -1 = a^2

2c - 1 = a^2

sqrt(2c - 1) = a

Now, if you do the same for "two-steps" and three "three steps" there comes a pattern.

sqrt(2c - 1) = a for one step sets.

sqrt(4c - 4) = a for two step sets.

sqrt (6c -9) = a for three step sets.

Do you see the pattern? Multiply by two for the left number, and the next number comes the square series. So, you can essentially predict the equation for any amount of step sets. If you have even read this far, I congratualate you.

Now, the next part is essential. Not just any number will work in a pythagorean set, so, formulaes are needed for this too.

For every formula that has an odd "right number," you go by halves of the left number to find a viable "a". This is hard to explain, because I don't know how to show it algebraically.

Example: The equation sqrt(2c - 1) = a has an odd right number. So, viable "a's" in this case are: 1,3,5,7,9 ect. A.K.A., any odd number.

For equations with even right numbers, any multiple of one half the left number with make a usable "a." Example: sqrt(4c -4) = a: 2, 4 ,6 , 8, 10 all work as "a". Once again I congratulate you.

To find a usable c is much more complex. There are two starting points:

Equations with odd right numbers-------------------------Even right numbers

one step: 2x^2 + 2x + 1 = c two step: x^2 + 4x +5 = c

three step: 6x^2 + 6x + 3 = c four step: 2x^2 + 8x + 10 = c

Where, of course, "x" is any positive integer (you can do zero, but the triplet will either contain a zero, or be off in some other way, but it technically still works). The pattern for the left column is multiplying by 3 and for the other side multiplying by 2.

Finally (congrats), one last piece of information.

I noticed that all the original equations {sqrt(2c -1) = a ect} can be rewritten another, more unifying way.

sqrt(2ec - e^2) = a where e = (c-b) for ALL pythagorean triplets. I was very excited about this until I realized you could plug in (c-b) for e and distribute, and it eventually all comes back to c^2 - b^2 = a^2, also known as the original problem.

Thanks for any comments.

I'm too exhausted (sorry, long day) to go through this completely, but just eyeballing it, it looks like some screwy things happen with the special case of a=0, b=0, c=0.

In order for the set 0,0,0 to be predicted in that equation series, you'd have to make a zero-step equation (0-0 = 0), which would be sqrt(c - 0) = a I imagine.

sqrt(0-0) = 0

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Oh, and I forgot to write the reason why I posted this. I was thinking that it might be useful to predict pythagorean triplets in a different way than what I read about, which seemed a bit too random to help. With these equations, you can define a length of a side of a triangle, and then find the corresponding triplet. That couldn't be done with the previous equation I saw.

Well first of all kudos, this is not new but to find it on one's own is remarkable I'd say. Now I have one question, by going back to the usual formulas for Pythagorean triples, one could probably show that your formula for c is right, however I guess you didn't find it that way, then how did you find the polynomials for c?
Again, very nice.

Well first of all kudos, this is not new but to find it on one's own is remarkable I'd say. Now I have one question, by going back to the usual formulas for Pythagorean triples, one could probably show that your formula for c is right, however I guess you didn't find it that way, then how did you find the polynomials for c?
Again, very nice.

Thanks for kudos. The polynominals I found for "c" were found simply by plugging in the next "a" in the formula, then graphing each successive "c". It was just a guess that there might be a pattern there, and I lucked out.

However, it's important to note that while I did prove most of that stuff, I didn't for the c polynominals, or the pattern that comes with them. I just noticed what was happening and jumped on it.

Fucking mathematicians. Always taking my ideas.

Thanks for kudos. The polynominals I found for "c" were found simply by plugging in the next "a" in the formula, then graphing each successive "c". It was just a guess that there might be a pattern there, and I lucked out.

However, it's important to note that while I did prove most of that stuff, I didn't for the c polynominals, or the pattern that comes with them. I just noticed what was happening and jumped on it.

Fucking mathematicians. Always taking my ideas.

I was looking at the formulas for Pyth triples and people only really care about primitive ones. But I did find a formula for the general imprimitive case here http://mathforum.org/dr.math/faq/faq.pythag.triples.html where it says Formula for all pythagorean triples. If you work with it a little, you get exactly what you have, including the polynomials for c. Also, careful what you say about mathematicians!.

Thanks for the link. I have rules for the primitive ones as well, at least, that applies to my little stchick here.

Primitives can only come from the two equations sqrt(2c -1) = a and sqrt(4c -4), with any odd number being okay for "a" in the first, and any odd multiple of 2 in the second. Anything equation 3-step and beyond won't work for primitives because you all triples are covered in the first two (because every even and every odd "a" can be used in the first two). But, whatever.

Thanks again for the link and the comments. That particular equation was the one I was shown by my math teacher in response to all this stuff. I didn't know how it was proven though, which was what I was really interested in, and now I do. (All this was done through a periodically through a month in math class instead of doing work, and so when he asked me why I wasn't doing my work, I showed him this in hopes I could salvage a detention or two.)