My most general result with sums of squares is that in general there must always exist nonzero x and y, such that for an integer n equal to 1 or higher, and an integer m equal to 3 or higher:

**x**

^{2}+ (m-1)y^{2}= m^{n}Where n starts at 1. Often I like to start it at zero so it's a count of iterations but is prettier starting at 1.

For a sum of c+1 squares:

**m = s**

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

**x**

^{2}+ (**s**

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

^{2}= m^{n}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

So for 5 squares, I'll need 4 s's and I'll use primes: 2, 3, 5, and 7

Then m = 4 + 9 + 25 + 49 + 1 = 88

1

^{2}+ 87*1

^{2}= 88

First iteration: (-86)

^{2}+ 87*(2)

^{2}= 88

^{2}

Which is: (-86)

^{2}+ 4*(2)

^{2}+ 9*(2)

^{2 }+ 25*(2)

^{2 }+ 49*(2)

^{2}= 88

^{2}

Which is:

**86**

^{2}+ 4^{2}+ 6^{2 }+ 10^{2 }+ 14^{2}= 88^{2}Second iteration: (-260)

^{2}+ 87*(-84)

^{2}= 88

^{3}

Which is: (-260)

^{2}+ 4*(-84)

^{2}+ 9*(-84)

^{2}+ 25*(-84)

^{2}+ 49*(-84)

^{2 }= 88

^{3}

Which is: 260

^{2}+ 168

^{2}+ 252

^{2}+ 420

^{2}+ 588

^{2 }= 88

^{3}

Third iteration: (7048)

^{2}+ 87*(-344)

^{2}= 88

^{4}

Which is: (7048)

^{2}+ 4*(-344)

^{2}+ 9*(-344)

^{2 }+ 25*(-344)

^{2 }+ 49*(-344)

^{2}= 88

^{4}

Which is:

**7048**

^{2}+ 688^{2}+ 1032^{2 }+ 1720^{2 }+ 2408^{2 }= 88^{4}First: 4

^{2}+ 6

^{2 }+ 10

^{2 }+ 14

^{2 }+ 86

^{2}= 88

^{2}

Second: 688

^{2}+ 1032

^{2 }+ 1720

^{2 }+ 2408

^{2 }+ 7048

^{2}= 88

^{4}

Where I can divide off 64 to get:

**86**

^{2}+ 129

^{2 }+ 215

^{2 }+ 301

^{2 }+ 881

^{2}= 64*11

^{4}

Which is: 86

^{2}+ 129

^{2 }+ 215

^{2 }+ 301

^{2 }+ 881

^{2}= 968

^{2}

So I found, two sums of fives squares that give a square:

**4**

^{2}+ 6^{2 }+ 10^{2 }+ 14^{2 }+**86**

^{2}**= 88**

^{2}**86**

^{2}+ 129^{2 }+ 215^{2 }+ 301^{2 }+ 881

^{2}**= 968**

^{2}Some things are worth noting. First I'm setting some factors of the squares, so the other thing is I can pick some factors. And for this example chose to use prime numbers.

But those can conceivably divide off at some iteration, or be divided off, as you can choose to divide off common factors or not. Oh yeah, signs can be switched at each iteration to get alternate values. I tend to just go with the flow in terms of signs. Makes it easier for me just typing up demonstration posts.

It is interesting that 86 popped up twice. There could be all kinds of patterns that emerge, but to go searching for them generating a lot more would be best so a computer program would be better.

Guess it could the kind of project some enterprising number theorist might do as who knows what could be learned when you can play at this scale so easily.

But you don't have to be a specialist, as the mathematics is so elementary. Really cool! You only need to know basic algebra and have the desire to explore.

James Harris