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So we’re all familiar with the famous Pythagorean Theorem for a right angled 2D triangles, which can be written as $x^2 + y^2 = r^2$ ($x$ is the base, $y$ is the height, $r$ is the hypotenuse).

There exists another formula for 3D triangles that can be written as $x^2 + y^2 + z^2 =r^2$.

Given the pattern here in dimensions, does there exist a 4D Pythagorean theorem, $x_1^2 + x_2^2 + x_3^2 + x_4^2 = r^2$? Is there a general Pythagorean theorem than applies for n dimensions, $x_1^2 +x_2^2 +x_3^2 + x_4^2 +…+x_n^2 =r^2$?

If so, how can we prove it if we can’t use any geometry?

And if not, for which dimensions does this Pythagorean pattern hold true?

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    I think the "theorem" is taken as the definition of length in higher dimensions. The distance from the origin to $(x_1,x_2,\dots,x_n)$ is defined to be $\sqrt{x_1^2+x_2^2+\cdots+x_n^2}$. – Gerry Myerson Sep 02 '22 at 01:19
  • The distance formula (calculating the square of a length from the squares of perpendicular lengths) certainly generalizes to $n$ dimensions. Perhaps more interestingly, de Gua's theorem in three dimensions relates squares of areas; this idea generalizes to $n$ dimensions, with the dimensions of the quantities involved increasing at each stage. (Lengths in $2$ dimensions; areas in $3$ dimensions; volumes in $4$ dimensions; etc, etc, etc.) – Blue Sep 02 '22 at 01:43
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    "[H]ow can we prove it if we can’t use any geometry?" This is a benefit of coordinate geometry (and linear algebra, etc). It lets us exploit algebra to explore geometry we can't "see". – Blue Sep 02 '22 at 01:48
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    @GerryMyerson but you can prove this inductively purely geometrically. So the "definition" is really a codification the the Euclidean $n$-dimensional result. – Thomas Andrews Sep 02 '22 at 02:03
  • Related: https://math.stackexchange.com/questions/1588798/how-would-pythagoreans-theorem-work-in-higher-dimensions-general-question – Baguette Boy Sep 03 '22 at 18:39

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