**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.

For example:

**495**

^{2}+ 4888^{2}= 17^{6}That result looks like something from antiquity, but it's something I just recently noticed which follows from my own basic math research where I could have had the result back in 2008.

That is the underlying math discovery was in 2008, but I finally decided to name the thing, and with that event figured I should play with it some more and finally noticed this sum of squares result. I have lots of math laying around. Every once in a while I look back over something old.

Underlying technique allows you to do calculations with half the bits. The calculations to get there always involve half as many bits, and maybe that could be useful to someone? I don't know. May as well mention it.

The math is easy too.

Found out I could get answers for:

**(x**

^{2}+ ay^{2})(u^{2}+ bv^{2}) = p^{2}+ cq^{2}Where I was web searching trying to see what mathematicians knew in this area--I am NOT a mathematician I should emphasize--and came across a search result which brought up a page where someone was asking that question and I answer it easily.

Here's an example:

**(x**

^{2}+ 2y^{2})(u^{2}+ 3v^{2}) = p^{2}+ 11q^{2}Where my methods generate a potential infinity of solutions.

One of the solutions is: x = -1, y = 2, u = -2, v = 2, p = -10, q = 2

Gave my approach at the site too as an answer to the poster's question and it looks like it was promptly deleted! Here's a link to the site with the question.

That actually bugged me a bit but then made me feel better, as I have other number theory results like my own prime counter which I discovered over a decade ago. It's clear I'm doing my part, so the failure is elsewhere.

Regardless, something with which to do things. Does it really matter if some mathematicians somewhere acknowledge it?

James Harris