**x**

^{2}+ (m-1)y^{2}= m^{n+1}I have m raised to n+1 so that n is a count of iterations. But you can start n at one, which I'll show, with a sum of exactly two squares, which is true in general if m-1 is a square.

**x**

^{2}+ y^{2}= m^{n}will always have nonzero integer solutions for x and y, as long as m - 1 is a square and n is greater than 0. Here n starts at 1, because I think it looks prettier and you have a sum of exactly two squares.

And the general principle is then obvious:

**m = s**

_{1}^{2}+...+s_{c}^{2}+ 1

**x**

^{2}+ (**s**

_{1}^{2}+...+s_{c}^{2}**)y**

^{2}= m^{n}Where c is the number of sums of squares desired minus one. For example for two squares, c = 1.

In general the BQD Iterator is:

Given nonzero integers u and v with

u

^{2}+ (s

_{1}

^{2}+...+ s

_{c}

^{2})v

^{2}= F

then it must also be true that

(u - (s

_{1}

^{2}+..+s

_{c}

^{2})v)

^{2}+ (s

_{1}

^{2}+...+s

_{c}

^{2})(u + v)

^{2}= (s

_{1}

^{2}+...+ s

_{c}

^{2}+ 1)*F

And starting with u = v = 1 or -1, is necessary so that the initial F equals m.

Notice with this result if you square the square which is the result of this sum of squares then that square is also a sum of squares which is you can generate with just two more iterations.

_{1}

^{2}+ s

_{2}

^{2}+ 1

x

^{2}+ (s

_{1}

^{2}+ s

_{2}

^{2})y

^{2}= m

^{n}

And the BQD Iterator is:

Given nonzero integers u and v with:

u

^{2}+ (s

_{1}

^{2}+ s

_{2}

^{2})v

^{2}= F

then it must also be true that

(u - (s

_{1}

^{2}+s

_{2}

^{2})v)

^{2}+ (s

_{1}

^{2}+s

_{2}

^{2})(u + v)

^{2}= (s

_{1}

^{2}+ s

_{2}

^{2}+ 1)*F

And I'll keep it simple with s

_{1}= 2, and s

_{2}= 3, so m = 4 + 9 + 1 = 14.

And I'll use u = v = 1, so F = 14, to start.

Then first iteration: (-12)

^{2}+ 13(2)

^{2}= (-12)

^{2}+ 4(2)

^{2}+ 9(2)

^{2}=14

^{2}

Second iteration: (-38)

^{2}+ 13(-10)

^{2}= (-38)

^{2}+ 4(-10)

^{2}+ 9(-10)

^{2}=14

^{3}

And you can keep going indefinitely. But I'll stop here, and write that last prettier:

**38**

^{2}+ 20^{2}+ 30^{2}=14^{3}And I have a sum of three squares with a simple method for finding them. You can vary signs at each iteration to find other solutions.

Oh yeah, I tend to go higher but you also always have a result with all squares from the first iteration, or of course with any odd iteration as for instance 14

^{4}is also a square, and writing the first iteration with my numerical example out prettier:

**12**

^{2}+ 4^{2}+ 6^{2}=14^{2}Interesting. You can divide 4 off, and order from least to greatest:

**2**

^{2}+ 3^{2}+ 6^{2}= 7^{2}Guess I shouldn't just ignore just a sum of squares. Though to me that's the Old School, which is well worked out already.

Am curious enough to see that third iteration.

Third iteration: (92)

^{2}+ 13(-48)

^{2}= (92)

^{2}+ 4(-48)

^{2}+ 9(-48)

^{2}=14

^{4}

And prettifying, as well as looking at as all squares:

**92**

^{2}+ 96^{2}+ 144^{2}= 196^{2}= (14^{2})^{2}Oh yeah, can divide off 16 here, and will as I'm curious:

**23**

^{2}+ 24^{2}+ 36^{2}= 49^{2}= (7^{2})^{2}Now that has a curious feature:

**23**

^{2}+ 24^{2}+ (6^{2})^{2}= (7^{2})^{2}No radicals needed. Integers rule.

That is MY research result! Yeah!!! Woo hoo! You can in essence just dial up whatever number of squares you wish! Just for the fun of it. Probably would be better to automate it with a computer program.

This result lets you put any positive integer you want on that last exponent.

Doing web searches as usual and found there are results with sums of three squares, including identities which include my examples. But that's just confirmation of the approach.

The advantage I see is the generality and simplicity of this route.

So I can just pick any arbitrary sum of squares if I wished and generate solutions. Could do a hundred if I wanted, but would definitely program a computer to do such a result!

James Harris

## No comments:

Post a Comment