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

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

Beginner2022-07-12Added 12 answers

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, $({\xi}_{1},{\xi}_{2},{\xi}_{3},{\xi}_{4})$ where ${\xi}_{1},{\xi}_{2},{\xi}_{3},{\xi}_{4}$ are rational numbers, and see if the line $(1,0,0,0)+t({\xi}_{1},{\xi}_{2},{\xi}_{3},{\xi}_{4})$ intersects S4 at another rational point.

$\begin{array}{rl}(1+t{\xi}_{1}{)}^{2}+{t}^{2}{\xi}_{2}^{2}+{t}^{2}{\xi}_{3}^{2}+{t}^{2}{\xi}_{4}^{1}& =1\\ 2t{\xi}_{1}+{t}^{2}({\xi}_{1}^{2}+{\xi}_{2}^{2}+{\xi}_{3}^{2}+{\xi}_{4}^{2})& =0\\ t& =-{\displaystyle \frac{2{\xi}_{1}}{{\xi}_{1}^{2}+{\xi}_{2}^{2}+{\xi}_{3}^{2}+{\xi}_{4}^{2}}}\end{array}$

So, for any four rational numbers ${\xi}_{1},{\xi}_{2},{\xi}_{3},{\xi}_{4}$

$({\displaystyle \frac{-{\xi}_{1}^{2}+{\xi}_{2}^{2}+{\xi}_{3}^{2}+{\xi}_{4}^{2}}{{\xi}_{1}^{2}+{\xi}_{2}^{2}+{\xi}_{3}^{2}+{\xi}_{4}^{2}}},-{\displaystyle \frac{2{\xi}_{1}{\xi}_{2}}{{\xi}_{1}^{2}+{\xi}_{2}^{2}+{\xi}_{3}^{2}+{\xi}_{4}^{2}}},-{\displaystyle \frac{2{\xi}_{1}{\xi}_{3}}{{\xi}_{1}^{2}+{\xi}_{2}^{2}+{\xi}_{3}^{2}+{\xi}_{4}^{2}}},-{\displaystyle \frac{2{\xi}_{1}{\xi}_{4}}{{\xi}_{1}^{2}+{\xi}_{2}^{2}+{\xi}_{3}^{2}+{\xi}_{4}^{2}}},)$

is a rational point on S4

$\begin{array}{rl}(1+t{\xi}_{1}{)}^{2}+{t}^{2}{\xi}_{2}^{2}+{t}^{2}{\xi}_{3}^{2}+{t}^{2}{\xi}_{4}^{1}& =1\\ 2t{\xi}_{1}+{t}^{2}({\xi}_{1}^{2}+{\xi}_{2}^{2}+{\xi}_{3}^{2}+{\xi}_{4}^{2})& =0\\ t& =-{\displaystyle \frac{2{\xi}_{1}}{{\xi}_{1}^{2}+{\xi}_{2}^{2}+{\xi}_{3}^{2}+{\xi}_{4}^{2}}}\end{array}$

So, for any four rational numbers ${\xi}_{1},{\xi}_{2},{\xi}_{3},{\xi}_{4}$

$({\displaystyle \frac{-{\xi}_{1}^{2}+{\xi}_{2}^{2}+{\xi}_{3}^{2}+{\xi}_{4}^{2}}{{\xi}_{1}^{2}+{\xi}_{2}^{2}+{\xi}_{3}^{2}+{\xi}_{4}^{2}}},-{\displaystyle \frac{2{\xi}_{1}{\xi}_{2}}{{\xi}_{1}^{2}+{\xi}_{2}^{2}+{\xi}_{3}^{2}+{\xi}_{4}^{2}}},-{\displaystyle \frac{2{\xi}_{1}{\xi}_{3}}{{\xi}_{1}^{2}+{\xi}_{2}^{2}+{\xi}_{3}^{2}+{\xi}_{4}^{2}}},-{\displaystyle \frac{2{\xi}_{1}{\xi}_{4}}{{\xi}_{1}^{2}+{\xi}_{2}^{2}+{\xi}_{3}^{2}+{\xi}_{4}^{2}}},)$

is a rational point on S4

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