Abstract
This article provides the geometric and algebraic proof of the variant equation of the Pythagorean theorem x^2-xy+y2=z^2 . The hypothesis that will be proven is that just as squares govern the original version x^2+y^2=z^2 , hexagons are found to govern x^2-xy+y^2=z^2 . Both the special case x=y  and general case of x≠y  are examined.
Highlights
The equation that governs the Pythagorean theorem x2 + y2 = z2 has been explored in-depth both in its geometric and algebraic significance, with many proofs emerging throughout the years (Euclid et al 1908, Maor 2007)
The cosine rule given by z2 = y2 + x2 − 2xycos(γ) [Pickover, 2012] suggests that just as a reference angle of γ = 90 degrees (where cos(90°) = 0 indicates a relation between squares with an external angle of 90 degrees) reducing the said equation to x2 + y2 = z2, a reference angle of γ = 60 degrees (where cos(60°) = 1⁄2 indicates a relation between hexagons with an external angle of 60 degrees) reducing the said equation to x2 − xy + y2 = z2
Hypothesis It is the hypothesis of this article that just as squares govern the area relation between the three sides of a right-angled triangle expressed as x2 + y2 = z2, hexagons govern the area relation between the three sides of a 60-degree-reference triangle expressed as x2 − xy + y2 = z2
Summary
The equation that governs the Pythagorean theorem x2 + y2 = z2 has been explored in-depth both in its geometric and algebraic significance, with many proofs emerging throughout the years (Euclid et al 1908, Maor 2007). One variation of this equation is x2 − xy + y2 = z2. Cases that are key to understanding the geometric and algebraic behavior of such equations are x = y and x ≠ y.
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