somemath.blogspot.com/2015/01/considering-product-of-sum-of-squares.html

And I realized the problem is actually another interesting one in its own right, so here is another BQD Iterator result.

Given is the Diophantine equation:

**(x**

^{2}+ ay^{2})(u^{2}+ bv^{2}) = p^{2}+ cq^{2}You can find an infinity of solutions as long as c = ab + a + b, using BQD Iterators.

Like with a = 2, b = 3, which gives c = 11:

**(x**

^{2}+ 2y^{2})(u^{2}+ 3v^{2}) = p^{2}+ 11q^{2}And to me that's really interesting in terms of compactness.

To see two of the the three iterators needed go to the post, but I'll note that you can get c easily by letting all other variables equal 1, as then you have:

(1 + a)(1 + b) = 1 + c

That using that value will work over infinity is the advance this research brings,and calculating x, y, u, v, p and q is easy with three BQD Iterators. For those who know my research, it's like, oh yeah, easy.

But notice that two products of squares can compress into another, so I'm thinking to myself that's worth noting. There is an asymmetry there I wouldn't have just assumed. Part of me for some reason isn't totally surprised, so I guess I'm building a mathematical intuition about these things.

There is a product of exponents raised to 2 on one side of the equals with a sum of them raised to 2 on the other, but that product is with independent variables. Still, I don't see myself considering this type of problem on my own, which shows the importance of looking at what others are doing.

Asymmetry rears up again as a norm. Interesting.

Clearly there is a special mathematics to these kind of sums of squares, which is rather compact. Worth noting in its own post.

Curious readers can try to solve the Diophantine equations with what they think they know to compare and contrast. Can you figure out c? How do you get answers? If you can, how easily can you do it? If you struggle, why?

To me mathematics should be not just what you think you know, but what you can learn that works best.

James Harris

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