kramberol

2022-07-11

pythagorean theorem extensions
are there for a given integer N solutions to the equations
$\sum _{n=1}^{N}{x}_{i}^{2}={z}^{2}$
for integers ${x}_{i}$ and zan easier equation given an integer number 'a' can be there solutions to the equation
$\sum _{n=1}^{N}{x}_{i}^{2}={a}^{2}$
for N=2 this is pythagorean theorem

Tanner Hamilton

Just to avoid notation bloat, let's let N=4 and let the implied generalization take care of the rest of the question. The question can be restated as asking whether there are rational points on the 4-sphere ${\mathbb{S}}_{4}:{x}_{1}^{2}+{x}_{2}^{2}+{x}_{3}^{2}+{x}_{4}^{2}=1$. We know that (1,0,0,0) is a (trivial) rational point on S4. From that point, we pick a rational direction, $\left({\xi }_{1},{\xi }_{2},{\xi }_{3},{\xi }_{4}\right)$ where ${\xi }_{1},{\xi }_{2},{\xi }_{3},{\xi }_{4}$ are rational numbers, and see if the line $\left(1,0,0,0\right)+t\left({\xi }_{1},{\xi }_{2},{\xi }_{3},{\xi }_{4}\right)$ intersects S4 at another rational point.
$\begin{array}{rl}\left(1+t{\xi }_{1}{\right)}^{2}+{t}^{2}{\xi }_{2}^{2}+{t}^{2}{\xi }_{3}^{2}+{t}^{2}{\xi }_{4}^{1}& =1\\ 2t{\xi }_{1}+{t}^{2}\left({\xi }_{1}^{2}+{\xi }_{2}^{2}+{\xi }_{3}^{2}+{\xi }_{4}^{2}\right)& =0\\ t& =-\frac{2{\xi }_{1}}{{\xi }_{1}^{2}+{\xi }_{2}^{2}+{\xi }_{3}^{2}+{\xi }_{4}^{2}}\end{array}$
So, for any four rational numbers ${\xi }_{1},{\xi }_{2},{\xi }_{3},{\xi }_{4}$
$\left(\frac{-{\xi }_{1}^{2}+{\xi }_{2}^{2}+{\xi }_{3}^{2}+{\xi }_{4}^{2}}{{\xi }_{1}^{2}+{\xi }_{2}^{2}+{\xi }_{3}^{2}+{\xi }_{4}^{2}},-\frac{2{\xi }_{1}{\xi }_{2}}{{\xi }_{1}^{2}+{\xi }_{2}^{2}+{\xi }_{3}^{2}+{\xi }_{4}^{2}},-\frac{2{\xi }_{1}{\xi }_{3}}{{\xi }_{1}^{2}+{\xi }_{2}^{2}+{\xi }_{3}^{2}+{\xi }_{4}^{2}},-\frac{2{\xi }_{1}{\xi }_{4}}{{\xi }_{1}^{2}+{\xi }_{2}^{2}+{\xi }_{3}^{2}+{\xi }_{4}^{2}},\right)$
is a rational point on S4

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