**5**

^{2}+ 20^{2}= 17*5^{2}**8**

^{2}+ 19^{2}= 17*5^{2}

^{}**13**

^{2}+ 16^{2}= 17*5^{2}These were found using what I decided to call a Binary Quadratic Diophantine iterator, or BQD Iterator for short, which lets you find in general solutions for x and y with:

**x**

^{2}+ (m-1)y^{2}= F*m^{n}When F = x

_{0}

^{2}+ (m-1)y

_{0}

^{2}, my research proves there are non-zero integer solutions for x and y, where n is a count of iterations.

To get my examples used x

_{0}= 1, and y

_{0}= 2, and picked m = 5 so that I had 4 which I could pull into the square, to get a sum of two squares.

At each iteration you get a split point, where you can go positive or negative, which means as you iterate you may generate extra solutions for the same sum.

For my example I had a duplicate in the second iteration which is why there are only 3 distinct solutions instead of 4.

And the 5 being squared in this example is not an accident, as that's the second iteration. The first iteration has 17*5.

So yeah I could have kept going, and would have had a maximum of 6 cases of two squares summing to equal 17*5

^{3}, but less if there were duplicates.

I have an earlier post which shows how the solutions were found.

And I have additional research using the BQD Iterator showing how you get a desired number of sums of squares. Which could mean for math hobbyists, maybe could be useful in generating magic squares of squares?

James Harris