[seqfan] Re: Fun little identity (to me anyway)

[avik roy]

This identity can be more generalized:

[(n+p)^2 + k][n^2 + k] = [n(n+p) + k]^2 + kp^2;

This identity can contribute to expressing the product of sums of two squares as a sum of two squares [if k is a square].

As an illustration, check:

(3^2 + 2^2)*(10^2 + 2^2)=34^2 + 14^2



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