**34967**

^{2}+ 752^{2}+ 1128^{2 }+ 1880^{2 }+ 2632^{2}+ 4136^{2}+ 4888^{2}= 35721^{2}_____________________________

My basic result used to get an arbitrary length of 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 7 squares, I'll need 6 s's and I'll use primes: 2, 3, 5, 7, 11 and 13

Then m = 4 + 9 + 25 + 49 + 121 + 169 + 1 = 378

1

^{2}+ 377*1

^{2}= 378

First iteration: (-376)

^{2}+ 377*(2)

^{2}= 378

^{2}

Second iteration: (-1130)

^{2}+ 377*(-374)^{2}= 378^{3}
Third iteration: (139868)

^{2}+ 377*(-1504)^{2}= 378^{4}
Which is: (139868)

^{2}+ 4*(-1504)^{2}+ 9*(-1504)^{2 }+ 25*(-1504)^{2 }+ 49*(-1504)^{2}+ 121*(-1504)^{2}+ 169*(-1504)^{2}= 378^{4}
Can divide both sides by 16, and get rid of negatives to get:

Which is: 34967

^{2}+ 752^{2}+ 9*(376)^{2 }+ 25*(376)^{2 }+ 49*(376)^{2}+ 121*(376)^{2}+ 169*(376)^{2}= 189^{4}
And now get final result where will show as all squares:

**34967**

^{2}+ 752^{2}+ 1128^{2 }+ 1880^{2 }+ 2632^{2}+ 4136^{2}+ 4888^{2}= 35721^{2}
Which is interesting to me, I think. Looks more impressive that way. Of course watch it get built maybe less impressive? But still is, a sum of seven squares to get a square.

Don't really see a practical use, so to me? Is just pure math.

James Harris