**1**

^{2}+ 9*1^{2}= 10**8**

^{2}+ 9*2^{2}= 10^{2}**26**

^{2}+ 9*6^{2}= 10^{3}**28**

^{2}+ 9*32^{2}= 10^{4}Which goes out to infinity, where I used a simple rule.

If: u

^{2}+ 9v

^{2}= 10

^{a}

Then: (u - 9v)

^{2}+ 9(u + v)

^{2}= 10

^{a+1}

Which is just using my BQD Iterator. And notice that u and v can be positive or negative, while I like to show positive as is easier and looks prettier, which allows some selectivity, which I used behind the scenes to get my series above. Some choices made it more boring.

Like yeah so if you use a = 4, u = 28, and v = -32, then next is:

**316**

^{2}+ 9*4^{2}= 10^{5}So yeah,

*every*power of 10 can be written as a sum of two squares.

And was fascinated a couple of years ago that the general result is, for an integer n equal to 0 or higher, and an integer m equal to 3 or higher:

**x**

^{2}+ (m-1)y^{2}= m^{n+1}I have m raised to n+1 so that n is a count of iterations. And if m-1 is a square then every power of m can be shown as the sum of two squares.

And talk it all out in this post. So I just used m = 10 above.

James Harris