Searching for c among the cubes (Posted on 2014-11-23)

	Submitted by broll
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(a^3-b^3)/(a-b)=c^4. The trick is to make c itself a difference of consecutive cubes: c = (n+1)^3-n^3 = (3n^2+3n+1). Then: a = (3n^2+2n)*(3n^2+3n+1) b = (2n+1)*(3n^2+3n+1) (a-b) = (3n^2-1)(3n^2+3n+1). Substituting: c^2 -(a-b) = (3n^2+3n+1)^2-(3n^2-1) (3n^2+3n+1) = (n+1)^3+(2n+1)^3, as required. Clearly,(a^3-b^3)/(a-b) = a^2+ab+b^2. But if so ((3n^2+2n)*(3n^2+3n+1))^2 +(3n^2+2n)*(3n^2+3n+1) (2n+1)*(3n^2+3n+1) + ((2n+1)*(3n^2+3n+1))^2 = (9n^4+18n^3+15n^2+6n+1)(3n^2+3n+1)^2 = (3n^2+3n+1)^4 = c^4