## Friday, April 24, 2009

Did you know that (42^{2} + 111^{2})(2^{2} + 5^{2}) = (471^{2} + 432^{2}) = (639^{2} + 12^{2})?

Neither did I until just a minute ago. It turns out that the product of two sums of two integer squares is also equal to two sets of sums of other integer squares, and it's easy to figure out what they are.

Leonardo Pisano Fibonacci showed this in Proposition 6 of "The Book of Squares" in the year 1225, and I have read that even Diophantus was aware of it some 1800 years ago, but it seems it was Leibnitz who first applied complex numbers to this problem. Complex numbers make this problem simple.

Take the general case,

(a

^{2}+ b^{2})*(c^{2}+ d^{2})which factors to the complex roots

= [(a + ib)(a - ib)]*[(c + id)(c - id)]

rearrange the factors to get

= [(a + ib)(c + id)]*[(a - ib)(c - id)]

which multiplies out to

= [ (ac - bd) + i(ad + bc) ] * [ (ac - bd) - i(ad + bc) ]

but since this is a complex conjugate, it simplifies to

= (ac - bd)

^{2}+ (ad + bc)^{2}That is, given any product of sum of squares, you can easily find what other sum of squares represents the answer.

Further, we can also re-order the factors to give

= [(a + ib)(c - id)]*[(a - ib)(c + id)]

= [ (ac + bd) + i(bc - ad) ] * [ (ac + bd) - i(bc - ad) ]

= (ac + bd)

^{2}+ (bc - ad)^{2}Which is yet another solution represented as a sum of squares!

Try it! (42

^{2}+ 111

^{2})(2

^{2}+ 5

^{2}) = (471

^{2}+ 432

^{2}) = (639

^{2}+ 12

^{2})

Burton MacKenZie www.burtonmackenzie.com

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